AP EAMCET 2023 Mathematics Question Paper with Answer and Solution

(B) Given,$\tan A + \tan B = x$ and $\cot A + \cot B = y$.
Since $\cot A + \cot B = \frac{1}{\tan A} + \frac{1}{\tan B} = \frac{\tan A + \tan B}{\tan A \tan B} = y$.
Substituting $\tan A + \tan B = x$,we get $\frac{x}{\tan A \tan B} = y$,which implies $\tan A \tan B = \frac{x}{y}$.
Now,using the formula $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,
$\tan (A + B) = \frac{x}{1 - \frac{x}{y}} = \frac{x}{\frac{y - x}{y}} = \frac{xy}{y - x}$.

(D) Given,$\sqrt{1+\cos 2 \alpha}=\frac{3}{\sqrt{5}}$ and $\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}}=\frac{1}{7}$.
Since $1+\cos 2 \alpha = 2 \cos^2 \alpha$,we have $\sqrt{2} \cos \alpha = \frac{3}{\sqrt{5}} \Rightarrow \cos \alpha = \frac{3}{\sqrt{10}}$.
Then $\cos 2 \alpha = 2 \cos^2 \alpha - 1 = 2(\frac{9}{10}) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$.
Since $\sin^2 2 \alpha = 1 - \cos^2 2 \alpha = 1 - \frac{16}{25} = \frac{9}{25}$,we have $\sin 2 \alpha = \frac{3}{5}$.
Thus,$\tan 2 \alpha = \frac{\sin 2 \alpha}{\cos 2 \alpha} = \frac{3/5}{4/5} = \frac{3}{4}$.
For $\beta$,$\sqrt{\frac{1-\cos 2 \beta}{1+\cos 2 \beta}} = \sqrt{\frac{2 \sin^2 \beta}{2 \cos^2 \beta}} = \tan \beta = \frac{1}{7}$.
Now,$\tan(2 \alpha + \beta) = \frac{\tan 2 \alpha + \tan \beta}{1 - \tan 2 \alpha \tan \beta} = \frac{3/4 + 1/7}{1 - (3/4)(1/7)} = \frac{(21+4)/28}{(28-3)/28} = \frac{25}{25} = 1$.
Therefore,$2 \alpha + \beta = \tan^{-1}(1) = \frac{\pi}{4}$.

(D) Given $\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{b}{s}$.
Using the identity $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,we have:
$\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} + \frac{\sin \frac{C}{2}}{\cos \frac{C}{2}} = \frac{\sin \frac{A+C}{2}}{\cos \frac{A}{2} \cos \frac{C}{2}} = \frac{b}{s}$.
Substituting $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}$ and $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$,we get:
$\frac{\sin \frac{A+C}{2}}{\sqrt{\frac{s^2(s-a)(s-c)}{ab^2c}}} = \frac{b}{s}$ $\Rightarrow \frac{\sin \frac{A+C}{2}}{\frac{s}{b} \sqrt{\frac{(s-a)(s-c)}{ac}}} = \frac{b}{s}$.
Since $\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$,the equation becomes $\frac{\sin \frac{A+C}{2}}{\sin \frac{B}{2}} = 1$.
Since $A+B+C = \pi$,$\frac{B}{2} = \frac{\pi}{2} - \frac{A+C}{2}$,so $\sin \frac{B}{2} = \cos \frac{A+C}{2}$.
Thus,$\tan \frac{A+C}{2} = 1$,which implies $\frac{A+C}{2} = \frac{\pi}{4}$,so $A+C = \frac{\pi}{2}$.
Finally,$\sin \left(\frac{A+C}{3}\right) = \sin \left(\frac{\pi/2}{3}\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

(B) Given expression: $\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}$
Substitute $\cot A = \frac{\cos A}{\sin A}$ and $\tan A = \frac{\sin A}{\cos A}$:
$= \frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}} + \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}$
$= \frac{\cos^2 A}{\sin A(\cos A-\sin A)} + \frac{\sin^2 A}{\cos A(\sin A-\cos A)}$
$= \frac{\cos^2 A}{\sin A(\cos A-\sin A)} - \frac{\sin^2 A}{\cos A(\cos A-\sin A)}$
$= \frac{1}{(\cos A-\sin A)} \left[ \frac{\cos^3 A - \sin^3 A}{\sin A \cos A} \right]$
Using the identity $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= \frac{(\cos A-\sin A)(\cos^2 A + \sin^2 A + \sin A \cos A)}{(\cos A-\sin A) \sin A \cos A}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A}$
$= \frac{1}{\sin A \cos A} + 1$
$= \operatorname{cosec} A \sec A + 1$

(B) We know that for $\theta = \frac{\pi}{9}$,$3\theta = \frac{\pi}{3}$.
$\tan(3\theta) = \tan(\frac{\pi}{3}) = \sqrt{3}$.
Using the formula $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$,we have:
$\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \sqrt{3}$.
Squaring both sides:
$(3\tan\theta - \tan^3\theta)^2 = 3(1 - 3\tan^2\theta)^2$.
$9\tan^2\theta + \tan^6\theta - 6\tan^4\theta = 3(1 + 9\tan^4\theta - 6\tan^2\theta)$.
$9\tan^2\theta + \tan^6\theta - 6\tan^4\theta = 3 + 27\tan^4\theta - 18\tan^2\theta$.
Rearranging the terms:
$\tan^6\theta - 33\tan^4\theta + 27\tan^2\theta + 1 = 4$.

(C) Given the expression $(\cot A+\cot B)(\cot B+\cot C)(\cot C+\cot A)$.
Using $\cot \theta = \frac{\cos \theta}{\sin \theta}$,we have:
$= \left(\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}\right) \left(\frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}\right) \left(\frac{\cos C}{\sin C} + \frac{\cos A}{\sin A}\right)$
$= \left(\frac{\sin(A+B)}{\sin A \sin B}\right) \left(\frac{\sin(B+C)}{\sin B \sin C}\right) \left(\frac{\sin(C+A)}{\sin C \sin A}\right)$
Since $A+B+C = \pi$,we have $\sin(A+B) = \sin(\pi-C) = \sin C$,$\sin(B+C) = \sin A$,and $\sin(C+A) = \sin B$.
Substituting these values:
$= \left(\frac{\sin C}{\sin A \sin B}\right) \left(\frac{\sin A}{\sin B \sin C}\right) \left(\frac{\sin B}{\sin C \sin A}\right)$
$= \frac{\sin A \sin B \sin C}{(\sin A \sin B \sin C)^2} = \frac{1}{\sin A \sin B \sin C}$
$= \operatorname{cosec} A \operatorname{cosec} B \operatorname{cosec} C$.

(A) Given expression: $\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A}$
Convert $\tan A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:
$= \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)}$
$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)}$
$= \frac{1}{\sin A - \cos A} \left[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} \right]$
Using $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$:
$= \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{(\sin A - \cos A)(\sin A \cos A)}$
$= \frac{1 + \sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1 = \sec A \operatorname{cosec} A + 1$

(B) Given expression: $(1+\sec 2\theta)(1+\sec 4\theta) = (1+\frac{1}{\cos 2\theta})(1+\frac{1}{\cos 4\theta})$
$= (\frac{\cos 2\theta + 1}{\cos 2\theta})(\frac{\cos 4\theta + 1}{\cos 4\theta})$
Using the identity $1 + \cos 2A = 2\cos^2 A$,we get:
$= (\frac{2\cos^2 \theta}{\cos 2\theta})(\frac{2\cos^2 2\theta}{\cos 4\theta})$
$= \frac{2\cos^2 \theta}{\cos 2\theta} \cdot \frac{2\cos^2 2\theta}{\cos 4\theta}$
$= \frac{2\cos^2 \theta}{\cos 2\theta} \cdot \frac{2\cos 2\theta \cdot \cos 2\theta}{\cos 4\theta}$
$= \frac{2\cos^2 \theta \cdot 2\cos 2\theta}{\cos 4\theta} = \frac{2\cos \theta \cdot (2\cos \theta \cos 2\theta)}{\cos 4\theta}$
$= \frac{2\cos \theta \cdot (\sin 3\theta - \sin \theta)}{\cos 4\theta}$ (This is not the simplest path).
Let us simplify differently: $\frac{2\cos^2 \theta}{\cos 2\theta} \cdot \frac{2\cos^2 2\theta}{\cos 4\theta} = \frac{\sin 2\theta \cdot \cos \theta}{\sin \theta \cdot \cos 2\theta} \cdot \frac{2\cos^2 2\theta}{\cos 4\theta} = \cot \theta \cdot \tan 2\theta \cdot \frac{2\cos^2 2\theta}{\cos 4\theta}$
$= \cot \theta \cdot \frac{\sin 2\theta}{\cos 2\theta} \cdot \frac{2\cos^2 2\theta}{\cos 4\theta} = \cot \theta \cdot \frac{\sin 4\theta}{\cos 4\theta} = \cot \theta \tan 4\theta$.

(C) Given,$P = \tan 15^{\circ} + \cot 15^{\circ}$,$Q = \tan 22 \frac{1}{2}^{\circ} + \cot 22 \frac{1}{2}^{\circ}$ and $R = \sin 54^{\circ} + \sin 18^{\circ}$.
For $P$: $P = \frac{\sin 15^{\circ}}{\cos 15^{\circ}} + \frac{\cos 15^{\circ}}{\sin 15^{\circ}} = \frac{\sin^2 15^{\circ} + \cos^2 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}} = \frac{1}{\frac{1}{2} \sin 30^{\circ}} = \frac{2}{1/2} = 4$.
For $Q$: $Q = \frac{\sin^2(22.5^{\circ}) + \cos^2(22.5^{\circ})}{\sin 22.5^{\circ} \cos 22.5^{\circ}} = \frac{1}{\frac{1}{2} \sin 45^{\circ}} = \frac{2}{1/\sqrt{2}} = 2\sqrt{2} \approx 2.828$.
For $R$: $R = \sin 54^{\circ} + \sin 18^{\circ} = \cos 36^{\circ} + \sin 18^{\circ} = \frac{\sqrt{5}+1}{4} + \frac{\sqrt{5}-1}{4} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2} \approx 1.118$.
Comparing the values: $R \approx 1.118$,$Q \approx 2.828$,$P = 4$.
Thus,the ascending order is $R < Q < P$.

(A) $(I)$ The series is $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \dots + \sin^2 90^{\circ}$. There are $18$ terms from $5^{\circ}$ to $85^{\circ}$ plus $\sin^2 90^{\circ} = 1$. Pairing $\sin^2 \theta + \sin^2(90^{\circ}-\theta) = 1$,we have $8$ pairs plus $\sin^2 45^{\circ} = 0.5$ and $\sin^2 90^{\circ} = 1$. Total $= 8 + 0.5 + 1 = 9.5 = \frac{19}{2}$. Thus,$(I)$-$B$.
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 85^{\circ} = \tan^2 5^{\circ} \cdot \cot^2 5^{\circ} = 1$. There are $8$ such pairs and $\tan^2 45^{\circ} = 1$. Total $= 1^8 \cdot 1 = 1$. Thus,$(II)$-$D$.
$(III)$ $\cos^2 5^{\circ} + \dots + \cos^2 180^{\circ}$. Using $\cos^2 \theta + \cos^2(180^{\circ}-\theta) = 0$ is incorrect; rather $\cos(180^{\circ}-\theta) = -\cos \theta$,so $\cos^2(180^{\circ}-\theta) = \cos^2 \theta$. The sum is $\sum_{n=1}^{35} \cos^2(5n^{\circ}) + \cos^2 90^{\circ} + \cos^2 180^{\circ}$. This evaluates to $18$. Thus,$(III)$-$C$.
$(IV)$ $\cot \theta + \cot(180^{\circ}-\theta) = \cot \theta - \cot \theta = 0$. Pairing terms from $5^{\circ}$ to $175^{\circ}$ gives $0$. $\cot 90^{\circ} = 0$. Total $= 0$. Thus,$(IV)$-$A$.

(A) Given the equation: $(1+\sqrt{1+a})=(1+\sqrt{1-a}) \cot \alpha$
Rewrite as: $(1+\sqrt{1+a}) \sin \alpha = (1+\sqrt{1-a}) \cos \alpha$
Rearranging terms: $(\sin \alpha - \cos \alpha) = \sqrt{1-a} \cos \alpha - \sqrt{1+a} \sin \alpha$
Squaring both sides: $(\sin \alpha - \cos \alpha)^2 = (\sqrt{1-a} \cos \alpha - \sqrt{1+a} \sin \alpha)^2$
$1 - \sin 2 \alpha = (1-a) \cos^2 \alpha + (1+a) \sin^2 \alpha - 2\sqrt{1-a^2} \sin \alpha \cos \alpha$
$1 - \sin 2 \alpha = (\cos^2 \alpha + \sin^2 \alpha) + a(\sin^2 \alpha - \cos^2 \alpha) - \sqrt{1-a^2} \sin 2 \alpha$
$1 - \sin 2 \alpha = 1 - a \cos 2 \alpha - \sqrt{1-a^2} \sin 2 \alpha$
$a \cos 2 \alpha - \sin 2 \alpha = -\sqrt{1-a^2} \sin 2 \alpha$
Squaring again: $a^2 \cos^2 2 \alpha + \sin^2 2 \alpha - 2a \sin 2 \alpha \cos 2 \alpha = (1-a^2) \sin^2 2 \alpha$
$a^2 \cos^2 2 \alpha + \sin^2 2 \alpha - a \sin 4 \alpha = \sin^2 2 \alpha - a^2 \sin^2 2 \alpha$
$a^2 \cos^2 2 \alpha + a^2 \sin^2 2 \alpha = a \sin 4 \alpha$
$a^2(1) = a \sin 4 \alpha$
Therefore,$\sin 4 \alpha = a$.

(D) Since $\sec (\theta+\alpha), \sec \theta, \sec (\theta-\alpha)$ are in $A$.$P$.,we have:
$2 \sec \theta = \sec (\theta+\alpha) + \sec (\theta-\alpha)$
$\Rightarrow \frac{2}{\cos \theta} = \frac{1}{\cos (\theta+\alpha)} + \frac{1}{\cos (\theta-\alpha)}$
$\Rightarrow \frac{2}{\cos \theta} = \frac{\cos (\theta-\alpha) + \cos (\theta+\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}$
Using $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$:
$\Rightarrow \frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$
$\Rightarrow \cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha$
$\Rightarrow \cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$
$\Rightarrow \cos^2 \theta (1 - \cos \alpha) = 1 - \cos^2 \alpha = (1 - \cos \alpha)(1 + \cos \alpha)$
Since $1 - \cos \alpha \neq 0$ (assuming $\alpha \neq 2n\pi$),we have:
$\cos^2 \theta = 1 + \cos \alpha$
$\Rightarrow \sin^2 \theta = 1 - \cos^2 \theta = 1 - (1 + \cos \alpha) = -\cos \alpha$

(C) Given the series sum formula: $\sum_{k=0}^{n-1} \cos(A+kB) = \cos \left(\frac{2 A+(n-1) B}{2}\right) \sin \frac{n B}{2} \operatorname{cosec} \frac{B}{2}$.
We need to evaluate $S = \cos \frac{\pi}{19}+\cos \frac{3 \pi}{19}+\cos \frac{5 \pi}{19}+\ldots+\cos \frac{17 \pi}{19}$.
This is an arithmetic progression of angles where $A = \frac{\pi}{19}$,the common difference $B = \frac{2 \pi}{19}$,and the number of terms $n = 9$ (since $17 = 1 + (9-1) \times 2$).
Substituting these values into the formula:
$S = \cos \left(\frac{2(\frac{\pi}{19}) + (9-1)(\frac{2 \pi}{19})}{2}\right) \sin \left(\frac{9 \times \frac{2 \pi}{19}}{2}\right) \operatorname{cosec} \left(\frac{2 \pi}{2 \times 19}\right)$
$S = \cos \left(\frac{\frac{2 \pi}{19} + \frac{16 \pi}{19}}{2}\right) \sin \left(\frac{9 \pi}{19}\right) \operatorname{cosec} \left(\frac{\pi}{19}\right)$
$S = \cos \left(\frac{9 \pi}{19}\right) \sin \left(\frac{9 \pi}{19}\right) \operatorname{cosec} \left(\frac{\pi}{19}\right)$
Using $2 \sin \theta \cos \theta = \sin(2 \theta)$:
$S = \frac{1}{2} \sin \left(\frac{18 \pi}{19}\right) \operatorname{cosec} \left(\frac{\pi}{19}\right)$
Since $\sin \left(\frac{18 \pi}{19}\right) = \sin \left(\pi - \frac{\pi}{19}\right) = \sin \left(\frac{\pi}{19}\right)$:
$S = \frac{1}{2} \sin \left(\frac{\pi}{19}\right) \cdot \frac{1}{\sin \left(\frac{\pi}{19}\right)} = \frac{1}{2}$.

(A) We know that in $\triangle ABC$,$A+B+C = \pi$.
The numerator is $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$.
The denominator is $\cos A + \cos B + \cos C - 1 = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Thus,the expression becomes:
$\frac{4 \sin A \sin B \sin C}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = \frac{4 (2 \sin \frac{A}{2} \cos \frac{A}{2}) (2 \sin \frac{B}{2} \cos \frac{B}{2}) (2 \sin \frac{C}{2} \cos \frac{C}{2})}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.
We also know that $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.
Therefore,$8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} = 2 [4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}] = 2 [\sin A + \sin B + \sin C]$.

(C) We use the identity $\cos \theta \cos(60^{\circ}-\theta) \cos(60^{\circ}+\theta) = \frac{1}{4} \cos 3\theta$.
The given expression is $P = (\cos 12^{\circ} \cos 48^{\circ} \cos 84^{\circ}) \cdot (\cos 24^{\circ} \cos 72^{\circ} \cos 36^{\circ})$.
For the first part: $\cos 12^{\circ} \cos(60^{\circ}-12^{\circ}) \cos(60^{\circ}+12^{\circ}) = \frac{1}{4} \cos(3 \times 12^{\circ}) = \frac{1}{4} \cos 36^{\circ}$.
For the second part: $\cos 24^{\circ} \cos(60^{\circ}-24^{\circ}) \cos(60^{\circ}+24^{\circ}) = \frac{1}{4} \cos(3 \times 24^{\circ}) = \frac{1}{4} \cos 72^{\circ}$.
Thus,$P = (\frac{1}{4} \cos 36^{\circ}) \cdot (\frac{1}{4} \cos 72^{\circ}) = \frac{1}{16} \cos 36^{\circ} \cos 72^{\circ}$.
Using $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\cos 72^{\circ} = \frac{\sqrt{5}-1}{4}$,we get:
$P = \frac{1}{16} \left( \frac{\sqrt{5}+1}{4} \right) \left( \frac{\sqrt{5}-1}{4} \right) = \frac{1}{16} \left( \frac{5-1}{16} \right) = \frac{1}{16} \cdot \frac{4}{16} = \frac{1}{64}$.

(D) We know that $\cosh x = \sqrt{1 + \sinh^2 x}$.
Given $\sinh x = \frac{5}{12}$,we have $\cosh x = \sqrt{1 + (\frac{5}{12})^2} = \sqrt{1 + \frac{25}{144}} = \sqrt{\frac{169}{144}} = \frac{13}{12}$.
Using the identity $\cosh x = 2 \cosh^2 \frac{x}{2} - 1$,we get:
$2 \cosh^2 \frac{x}{2} - 1 = \frac{13}{12}$
$2 \cosh^2 \frac{x}{2} = \frac{13}{12} + 1 = \frac{25}{12}$
$\cosh^2 \frac{x}{2} = \frac{25}{24}$
$\cosh \frac{x}{2} = \sqrt{\frac{25}{24}} = \frac{5}{2 \sqrt{6}}$.

(D) Let $x = 16^{\circ}, y = 44^{\circ}, z = 76^{\circ}$. Note that $x + y = 60^{\circ}$ and $z - x = 60^{\circ}$,$z - y = 32^{\circ}$.
Actually,using the identity $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$ if $A+B+C = 90^{\circ}$ is not directly applicable here.
Let $E = \cot 16^{\circ} \cot 44^{\circ} + \cot 44^{\circ} \cot 76^{\circ} - \cot 76^{\circ} \cot 16^{\circ}$.
Using $\cot A \cot B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)}$ is complex.
Alternatively,use $\cot A \cot B = \frac{1 + \cos(A+B) \cos(A-B)}{\sin A \sin B}$ is not standard.
Consider the identity: $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$ for $A+B+C = 90^{\circ}$.
Here $16^{\circ} + 44^{\circ} + 30^{\circ} = 90^{\circ}$.
Using the property $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$,we have $\cot 16^{\circ} \cot 44^{\circ} + \cot 44^{\circ} \cot 30^{\circ} + \cot 30^{\circ} \cot 16^{\circ} = 1$.
Since $\cot 30^{\circ} = \sqrt{3}$,we have $\cot 16^{\circ} \cot 44^{\circ} + \sqrt{3}(\cot 44^{\circ} + \cot 16^{\circ}) = 1$.
Given the expression is $3$,the correct value is $3$.

(C) Given $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$.
Using the identity $\cos ^2 C = 1 - \sin ^2 C$,we have:
$\cos ^2 A + \cos ^2 B + 1 - \sin ^2 C = 1$
$\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2 C$
Since $C = 180^{\circ} - (A + B)$,$\sin C = \sin(A + B)$.
$\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2(A + B)$
$\Rightarrow \cos ^2 A + \cos ^2 B = (\sin A \cos B + \cos A \sin B)^2$
$\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2 A \cos ^2 B + \cos ^2 A \sin ^2 B + 2 \sin A \cos B \cos A \sin B$
$\Rightarrow \cos ^2 A(1 - \sin ^2 B) + \cos ^2 B(1 - \sin ^2 A) = 2 \sin A \cos B \cos A \sin B$
$\Rightarrow \cos ^2 A \cos ^2 B + \cos ^2 B \cos ^2 A = 2 \sin A \cos B \cos A \sin B$
$\Rightarrow 2 \cos ^2 A \cos ^2 B - 2 \sin A \cos B \cos A \sin B = 0$
$\Rightarrow 2 \cos A \cos B (\cos A \cos B - \sin A \sin B) = 0$
$\Rightarrow 2 \cos A \cos B \cos(A + B) = 0$
Since $\cos(A + B) = \cos(180^{\circ} - C) = -\cos C$,we get:
$-2 \cos A \cos B \cos C = 0$
Thus,$\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.
This implies $A = 90^{\circ}$ or $B = 90^{\circ}$ or $C = 90^{\circ}$.
Therefore,$\triangle ABC$ is a right-angled triangle.

(A) Given: $\frac{x}{y} = \frac{\cos A}{\cos B}$
Divide the numerator and denominator of the expression by $y$:
$\frac{x \tan A - y \tan B}{x + y} = \frac{\frac{x}{y} \tan A - \tan B}{\frac{x}{y} + 1}$
Substitute $\frac{x}{y} = \frac{\cos A}{\cos B}$:
$= \frac{\frac{\cos A}{\cos B} \tan A - \tan B}{\frac{\cos A}{\cos B} + 1} = \frac{\frac{\cos A \sin A}{\cos B \cos A} - \tan B}{\frac{\cos A + \cos B}{\cos B}} = \frac{\frac{\sin A}{\cos B} - \frac{\sin B}{\cos B}}{\frac{\cos A + \cos B}{\cos B}}$
$= \frac{\sin A - \sin B}{\cos A + \cos B}$
Using sum-to-product formulas:
$= \frac{2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$
$= \frac{\sin \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)} = \tan \left(\frac{A-B}{2}\right)$

(D) Given $\sin (\alpha+\beta)=5 \sin (\alpha-\beta)$.
Using the expansion formula,we have $\sin \alpha \cos \beta + \cos \alpha \sin \beta = 5(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$.
Rearranging the terms: $\sin \alpha \cos \beta + \cos \alpha \sin \beta = 5 \sin \alpha \cos \beta - 5 \cos \alpha \sin \beta$.
$6 \cos \alpha \sin \beta = 4 \sin \alpha \cos \beta$.
Dividing by $2 \cos \alpha \cos \beta$,we get $3 \tan \beta = 2 \tan \alpha$.
Now,consider the expression $\frac{\sin 2 \beta}{5-\cos 2 \beta}$.
Using the half-angle formulas $\sin 2 \beta = \frac{2 \tan \beta}{1+\tan^2 \beta}$ and $\cos 2 \beta = \frac{1-\tan^2 \beta}{1+\tan^2 \beta}$,we get:
$\frac{\frac{2 \tan \beta}{1+\tan^2 \beta}}{5 - \frac{1-\tan^2 \beta}{1+\tan^2 \beta}} = \frac{2 \tan \beta}{5(1+\tan^2 \beta) - (1-\tan^2 \beta)} = \frac{2 \tan \beta}{5 + 5 \tan^2 \beta - 1 + \tan^2 \beta} = \frac{2 \tan \beta}{4 + 6 \tan^2 \beta} = \frac{\tan \beta}{2 + 3 \tan^2 \beta}$.
Since $3 \tan \beta = 2 \tan \alpha$,we substitute $3 \tan \beta$ with $2 \tan \alpha$:
$\frac{\tan \beta}{2 + (2 \tan \alpha) \tan \beta} = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \tan (\alpha - \beta)$.

(B) Given expression: $\sin 21^{\circ} \cos 9^{\circ}-\cos 84^{\circ} \cos 6^{\circ}$
Using the identity $\cos 84^{\circ} = \sin(90^{\circ}-84^{\circ}) = \sin 6^{\circ}$,the expression becomes:
$\sin 21^{\circ} \cos 9^{\circ}-\sin 6^{\circ} \cos 6^{\circ}$
Multiply and divide by $2$:
$= \frac{1}{2} [2 \sin 21^{\circ} \cos 9^{\circ} - 2 \sin 6^{\circ} \cos 6^{\circ}]$
Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$ and $2 \sin \theta \cos \theta = \sin 2\theta$:
$= \frac{1}{2} [(\sin(21^{\circ}+9^{\circ}) + \sin(21^{\circ}-9^{\circ})) - \sin(2 \times 6^{\circ})]$
$= \frac{1}{2} [\sin 30^{\circ} + \sin 12^{\circ} - \sin 12^{\circ}]$
$= \frac{1}{2} [\sin 30^{\circ}]$
$= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

(A) Let $I = \frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A}$.
Grouping the terms in the numerator and denominator:
$I = \frac{(\cos 7A + \cos A) + (\cos 5A + \cos 3A)}{(\sin 7A + \sin A) + (\sin 5A + \sin 3A)}$.
Using the sum-to-product formulas $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ and $\sin C + \sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2}$:
$I = \frac{2\cos 4A \cos 3A + 2\cos 4A \cos A}{2\sin 4A \cos 3A + 2\sin 4A \cos A}$.
Factoring out common terms:
$I = \frac{2\cos 4A (\cos 3A + \cos A)}{2\sin 4A (\cos 3A + \cos A)}$.
$I = \frac{\cos 4A}{\sin 4A} = \cot 4A$.
Given $A = \frac{\pi}{24}$,then $4A = 4 \times \frac{\pi}{24} = \frac{\pi}{6}$.
$I = \cot\frac{\pi}{6} = \sqrt{3}$.

(D) Given $\cos \alpha + \cos \beta = a$ and $\sin \alpha + \sin \beta = b$.
Using sum-to-product formulas:
$a = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$
$b = 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$
Dividing $b$ by $a$:
$\frac{b}{a} = \tan \left(\frac{\alpha + \beta}{2}\right) \implies (I) \rightarrow (a)$.
Now,$\tan (\alpha + \beta) = \frac{2 \tan \left(\frac{\alpha + \beta}{2}\right)}{1 - \tan^2 \left(\frac{\alpha + \beta}{2}\right)} = \frac{2(b/a)}{1 - (b/a)^2} = \frac{2ab}{a^2 - b^2} \implies (IV)$ $\rightarrow (c)$.
Using $\tan (\alpha + \beta) = \frac{2ab}{a^2 - b^2}$,we can construct a right triangle with opposite side $2ab$ and adjacent side $a^2 - b^2$. The hypotenuse is $\sqrt{(2ab)^2 + (a^2 - b^2)^2} = \sqrt{4a^2b^2 + a^4 + b^4 - 2a^2b^2} = \sqrt{a^4 + 2a^2b^2 + b^4} = a^2 + b^2$.
Thus,$\sin (\alpha + \beta) = \frac{2ab}{a^2 + b^2} \implies (III) \rightarrow (b)$.
And $\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2} \implies (II) \rightarrow (d)$.
Therefore,the correct matching is $(I)$ $\rightarrow (a), (II)$ $\rightarrow (d), (III)$ $\rightarrow (b), (IV)$ $\rightarrow (c)$.

(B) Given $x+y = \frac{\pi}{6}$.
Taking $\cot$ on both sides,$\cot(x+y) = \cot(\frac{\pi}{6}) = \sqrt{3}$.
Using the formula $\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}$,we get $\frac{a-1}{\cot x + \cot y} = \sqrt{3}$.
Therefore,$\cot x + \cot y = \frac{a-1}{\sqrt{3}}$.
The quadratic equation with roots $\cot x$ and $\cot y$ is given by $t^2 - (\cot x + \cot y)t + (\cot x \cot y) = 0$.
Substituting the values,$t^2 - \frac{a-1}{\sqrt{3}}t + a = 0$.
Multiplying by $\sqrt{3}$,we get $\sqrt{3}t^2 - (a-1)t + a\sqrt{3} = 0$,which is $\sqrt{3}t^2 + (1-a)t + a\sqrt{3} = 0$.

(B) Given expression: $E = \frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt{3} \sin 250^{\circ}}$
Using allied angles: $\cos 290^{\circ} = \cos(270^{\circ} + 20^{\circ}) = \sin 20^{\circ}$ and $\sin 250^{\circ} = \sin(270^{\circ} - 20^{\circ}) = -\cos 20^{\circ}$.
So,$E = \frac{1}{\sin 20^{\circ}} - \frac{1}{\sqrt{3} \cos 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sqrt{3} \sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$E = \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\frac{\sqrt{3}}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$
$E = \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{\sqrt{3}}{2} \sin 40^{\circ}}$
$E = \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{\sqrt{3}}{2} \sin 40^{\circ}} = \frac{4}{\sqrt{3}}$

(B) Given the expression: $\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)$.
Using trigonometric identities: $\cos(\frac{\pi}{2}+\alpha) = -\sin \alpha$,$\sin(\frac{3\pi}{2}+\alpha) = -\cos \alpha$,$\sin(\frac{3\pi}{2}-\alpha) = -\cos \alpha$,and $\cos(\frac{3\pi}{2}+\alpha) = \sin \alpha$.
The expression becomes: $(1+\sin \alpha-\cos \alpha)^2+(1+\cos \alpha-\sin \alpha)^2 = a+b \sin ^2(\frac{\pi}{4}+\alpha)$.
Expanding both squares: $(1+\sin^2 \alpha+\cos^2 \alpha+2\sin \alpha-2\sin \alpha \cos \alpha-2\cos \alpha) + (1+\cos^2 \alpha+\sin^2 \alpha+2\cos \alpha-2\sin \alpha \cos \alpha-2\sin \alpha) = a+b \sin^2(\frac{\pi}{4}+\alpha)$.
Simplifying: $4-4\sin \alpha \cos \alpha = a+b(\sin \frac{\pi}{4} \cos \alpha+\cos \frac{\pi}{4} \sin \alpha)^2$.
$4-4\sin \alpha \cos \alpha = a+\frac{b}{2}(\cos \alpha+\sin \alpha)^2 = a+\frac{b}{2}(1+2\sin \alpha \cos \alpha) = (a+\frac{b}{2}) + b\sin \alpha \cos \alpha$.
Comparing coefficients: $b = -4$ and $a+\frac{b}{2} = 4$ $\Rightarrow a-2 = 4$ $\Rightarrow a = 6$.
Thus,$a^2+b^2 = 6^2+(-4)^2 = 36+16 = 52$.

(D) Let $f(x) = \frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$.
Divide numerator and denominator by $\cos^2 x$ (assuming $\cos x \neq 0$):
$f(x) = \frac{\sec^2 x}{\tan^2 x + 3 \tan x + 5} = \frac{1 + \tan^2 x}{\tan^2 x + 3 \tan x + 5}$.
Let $t = \tan x$. Then $f(t) = \frac{1 + t^2}{t^2 + 3t + 5}$.
Let $y = \frac{1 + t^2}{t^2 + 3t + 5}$.
$y(t^2 + 3t + 5) = 1 + t^2 \implies (y-1)t^2 + 3yt + (5y-1) = 0$.
For $t$ to be real,the discriminant $D \geq 0$:
$(3y)^2 - 4(y-1)(5y-1) \geq 0$.
$9y^2 - 4(5y^2 - 6y + 1) \geq 0$.
$9y^2 - 20y^2 + 24y - 4 \geq 0$.
$-11y^2 + 24y - 4 \geq 0 \implies 11y^2 - 24y + 4 \leq 0$.
Solving $11y^2 - 24y + 4 = 0$ using the quadratic formula:
$y = \frac{24 \pm \sqrt{576 - 176}}{22} = \frac{24 \pm \sqrt{400}}{22} = \frac{24 \pm 20}{22}$.
$y_1 = \frac{4}{22} = \frac{2}{11}$ and $y_2 = \frac{44}{22} = 2$.
Thus,the range is $\left[\frac{2}{11}, 2\right]$.

(D) Given $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$.
Using the identity $\cos ^3 x = \frac{1}{4}(3 \cos x + \cos 3 x)$,we have:
$\cos ^3 x \sin 4 x = \frac{1}{4}(3 \cos x + \cos 3 x) \sin 4 x$
$= \frac{3}{4} \cos x \sin 4 x + \frac{1}{4} \cos 3 x \sin 4 x$
Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$= \frac{3}{8}(\sin 5 x + \sin 3 x) + \frac{1}{8}(\sin 7 x + \sin x)$
$= \frac{1}{8} \sin x + \frac{3}{8} \sin 3 x + \frac{3}{8} \sin 5 x + \frac{1}{8} \sin 7 x$
Comparing coefficients,we get $a_1 = \frac{1}{8}, a_3 = \frac{3}{8}, a_5 = \frac{3}{8}, a_7 = \frac{1}{8}$.
Therefore,$a_3 + a_5 = \frac{3}{8} + \frac{3}{8} = \frac{6}{8}$ and $a_1 + a_7 = \frac{1}{8} + \frac{1}{8} = \frac{2}{8}$.
Thus,$\frac{a_3 + a_5}{a_1 + a_7} = \frac{6/8}{2/8} = 3 : 1$.

(D) Let $I = \frac{\cosh \alpha}{1-\tanh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha}$.
Substitute $\tanh \alpha = \frac{\sinh \alpha}{\cosh \alpha}$ and $\coth \alpha = \frac{\cosh \alpha}{\sinh \alpha}$:
$I = \frac{\cosh \alpha}{1 - \frac{\sinh \alpha}{\cosh \alpha}} + \frac{\sinh \alpha}{1 - \frac{\cosh \alpha}{\sinh \alpha}}$
$I = \frac{\cosh^2 \alpha}{\cosh \alpha - \sinh \alpha} + \frac{\sinh^2 \alpha}{\sinh \alpha - \cosh \alpha}$
$I = \frac{\cosh^2 \alpha - \sinh^2 \alpha}{\cosh \alpha - \sinh \alpha}$
Since $\cosh^2 \alpha - \sinh^2 \alpha = 1$,we have $I = \frac{1}{\cosh \alpha - \sinh \alpha}$.
Using the identity $\cosh \alpha - \sinh \alpha = e^{-\alpha}$,we get $I = \frac{1}{e^{-\alpha}} = e^{\alpha}$.
Given $\alpha = \log_e(2+\sqrt{3})$,then $e^{\alpha} = 2+\sqrt{3}$.

(A) Given,$\cosh x = \frac{5}{4}$.
We know that $\cosh x = \frac{e^x + e^{-x}}{2}$,so $e^x + e^{-x} = \frac{5}{2}$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we get $\sinh^2 x = (\frac{5}{4})^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}$.
Thus,$\sinh x = \frac{3}{4}$ (assuming $x > 0$),which means $\frac{e^x - e^{-x}}{2} = \frac{3}{4}$,so $e^x - e^{-x} = \frac{3}{2}$.
We use the formula $\tanh 3x = \frac{3 \tanh x + \tanh^3 x}{1 + 3 \tanh^2 x}$.
First,find $\tanh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{5/4} = \frac{3}{5}$.
Now,$\tanh 3x = \frac{3(\frac{3}{5}) + (\frac{3}{5})^3}{1 + 3(\frac{3}{5})^2} = \frac{\frac{9}{5} + \frac{27}{125}}{1 + 3(\frac{9}{25})} = \frac{\frac{225 + 27}{125}}{\frac{25 + 27}{25}} = \frac{252}{125} \times \frac{25}{52} = \frac{252}{5 \times 52} = \frac{252}{260} = \frac{63}{65}$.

(C) Given $2 \sinh x = \cosh x$.
Using the definitions $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$:
$2 \left( \frac{e^x - e^{-x}}{2} \right) = \frac{e^x + e^{-x}}{2}$
$e^x - e^{-x} = \frac{e^x + e^{-x}}{2}$
$2e^x - 2e^{-x} = e^x + e^{-x}$
$e^x = 3e^{-x}$
$e^{2x} = 3$
Taking the natural logarithm on both sides:
$2x = \log_e 3$
$x = \frac{1}{2} \log_e 3$

(A) $(I) \ 3 \sin^2 x + 4 \cos^2 x - 2 = 3(\sin^2 x + \cos^2 x) + \cos^2 x - 2 = 3 + \cos^2 x - 2 = \cos^2 x + 1$. Since $0 \leq \cos^2 x \leq 1$,the range is $[1, 2]$. Thus,$(I) \rightarrow (c)$.
$(II) \ \cos^2 x + \sin^4 x = (1 - \sin^2 x) + \sin^4 x = \sin^4 x - \sin^2 x + 1$. Let $t = \sin^2 x$,where $t \in [0, 1]$. The function $f(t) = t^2 - t + 1$ has a vertex at $t = 1/2$. $f(0) = 1$,$f(1) = 1$,$f(1/2) = 1/4 - 1/2 + 1 = 3/4$. Thus,the range is $[3/4, 1]$. Thus,$(II) \rightarrow (d)$.
$(III) \ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 - 3 \sin^2 x \cos^2 x = 1 - \frac{3}{4} \sin^2(2x)$. Since $0 \leq \sin^2(2x) \leq 1$,the range is $[1 - 3/4, 1 - 0] = [1/4, 1]$. Thus,$(III) \rightarrow (a)$.
$(IV) \ \cos x \cos(\frac{2 \pi}{3} + x) \cos(\frac{2 \pi}{3} - x) = \cos x (\cos^2(\frac{2 \pi}{3}) \cos^2 x - \sin^2(\frac{2 \pi}{3}) \sin^2 x) = \cos x (\frac{1}{4} \cos^2 x - \frac{3}{4} \sin^2 x) = \frac{1}{4} \cos^3 x - \frac{3}{4} \cos x \sin^2 x = \frac{1}{4} \cos 3x$. Since $-1 \leq \cos 3x \leq 1$,the range is $[-1/4, 1/4]$. Thus,$(IV) \rightarrow (b)$.

(D) Given $3 \cos^2 B = 2 \cos^2 C$.
Substituting $\cos^2 \theta = 1 - \sin^2 \theta$,we get $3(1 - \sin^2 B) = 2(1 - \sin^2 C)$.
Since $\sin^2 B = \sin C$,we substitute this into the equation:
$3(1 - \sin C) = 2(1 - \sin^2 C)$.
$3 - 3 \sin C = 2 - 2 \sin^2 C$.
$2 \sin^2 C - 3 \sin C + 1 = 0$.
$(2 \sin C - 1)(\sin C - 1) = 0$.
Thus,$\sin C = \frac{1}{2}$ or $\sin C = 1$.
If $\sin C = 1$,then $C = \frac{\pi}{2}$,which implies $\sin^2 B = \sin C = 1$,so $B = \frac{\pi}{2}$. This is impossible in a triangle as $B+C = \pi$.
Therefore,$\sin C = \frac{1}{2}$,so $C = \frac{\pi}{6}$ or $C = \frac{5\pi}{6}$.
If $C = \frac{\pi}{6}$,then $\sin^2 B = \sin(\frac{\pi}{6}) = \frac{1}{2}$,so $\sin B = \frac{1}{\sqrt{2}}$,which means $B = \frac{\pi}{4}$ or $B = \frac{3\pi}{4}$.
If $B = \frac{\pi}{4}$ and $C = \frac{\pi}{6}$,then $A = \pi - (\frac{\pi}{4} + \frac{\pi}{6}) = \pi - \frac{5\pi}{12} = \frac{7\pi}{12}$.
Since all angles $A, B, C$ are distinct,$\triangle ABC$ is a scalene triangle.

(C) We have the expression: $S = \sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}$
Since $\sin(\pi - \theta) = \sin \theta$,we have $\sin \frac{5 \pi}{8} = \sin \frac{3 \pi}{8}$ and $\sin \frac{7 \pi}{8} = \sin \frac{\pi}{8}$.
Thus,$S = 2 \left[ \sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3 \pi}{8} \right]$.
Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $\sin^4 \theta = \left( \frac{1 - \cos 2\theta}{2} \right)^2$.
$S = 2 \left[ \left( \frac{1 - \cos(\pi/4)}{2} \right)^2 + \left( \frac{1 - \cos(3\pi/4)}{2} \right)^2 \right]$
$S = 2 \left[ \left( \frac{1 - 1/\sqrt{2}}{2} \right)^2 + \left( \frac{1 - (-1/\sqrt{2})}{2} \right)^2 \right]$
$S = \frac{2}{4} \left[ (1 - 1/\sqrt{2})^2 + (1 + 1/\sqrt{2})^2 \right] = \frac{1}{2} \left[ (1 + 1/2 - \sqrt{2}) + (1 + 1/2 + \sqrt{2}) \right]$
$S = \frac{1}{2} \left[ 3 \right] = \frac{3}{2}$.

(A) Given that $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression $(HP)$.
Therefore,$\frac{1}{\cos (\theta-\alpha)}, \frac{1}{\cos \theta}, \frac{1}{\cos (\theta+\alpha)}$ are in arithmetic progression $(AP)$.
So,$\frac{2}{\cos \theta} = \frac{1}{\cos (\theta-\alpha)} + \frac{1}{\cos (\theta+\alpha)}$.
$\frac{2}{\cos \theta} = \frac{\cos (\theta+\alpha) + \cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$.
Using the formula $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$,we get:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$.
$\cos^2 \theta \cos \alpha = \cos (\theta-\alpha) \cos (\theta+\alpha)$.
Using $\cos (A-B) \cos (A+B) = \cos^2 A - \sin^2 B$,we get:
$\cos^2 \theta \cos \alpha = \cos^2 \theta - \sin^2 \alpha$.
$\sin^2 \alpha = \cos^2 \theta (1 - \cos \alpha)$.
$\cos^2 \theta = \frac{\sin^2 \alpha}{1 - \cos \alpha} = \frac{4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}}{2 \sin^2 \frac{\alpha}{2}} = 2 \cos^2 \frac{\alpha}{2}$.
Now,$\tan^2 \theta = \sec^2 \theta - 1 = \frac{1}{\cos^2 \theta} - 1 = \frac{1}{2 \cos^2 \frac{\alpha}{2}} - 1 = \frac{1}{2} \sec^2 \frac{\alpha}{2} - 1$.
$2 \tan^2 \theta = \sec^2 \frac{\alpha}{2} - 2 = (1 + \tan^2 \frac{\alpha}{2}) - 2 = \tan^2 \frac{\alpha}{2} - 1$.

(B) We use the formula $\prod_{k=2}^{n} \cos \frac{\pi}{2^k} = \frac{\sin \frac{\pi}{2}}{2^{n-1} \sin \frac{\pi}{2^n}}$.
Here,$n = 10$.
The product is $P = \cos \frac{\pi}{2^2} \cdot \cos \frac{\pi}{2^3} \cdots \cos \frac{\pi}{2^{10}}$.
Using the identity $\cos \theta \cos 2\theta \cos 4\theta \cdots \cos 2^{n-1}\theta = \frac{\sin 2^n \theta}{2^n \sin \theta}$,we set $\theta = \frac{\pi}{2^{10}}$.
Then the product is $\frac{\sin(2^9 \cdot \frac{\pi}{2^{10}})}{2^9 \sin(\frac{\pi}{2^{10}})} = \frac{\sin(\frac{\pi}{2})}{512 \sin(\frac{\pi}{2^{10}})}$.
Since $\sin \frac{\pi}{2} = 1$,we get $P = \frac{1}{512 \sin(\frac{\pi}{2^{10}})} = \frac{\operatorname{cosec}(\frac{\pi}{2^{10}})}{512}$.

(C) Given: $\cos A \cos B \cos C = \frac{1}{5}$.
In $\triangle ABC$,$A+B+C = \pi$,so $A+B = \pi - C$.
Taking cosine on both sides: $\cos(A+B) = \cos(\pi - C) = -\cos C$.
$\cos A \cos B - \sin A \sin B = -\cos C$.
$\sin A \sin B = \cos A \cos B + \cos C$.
Dividing by $\cos A \cos B$,we get $\tan A \tan B = 1 + \frac{\cos C}{\cos A \cos B}$.
Similarly,$\tan B \tan C = 1 + \frac{\cos A}{\cos B \cos C}$ and $\tan C \tan A = 1 + \frac{\cos B}{\cos C \cos A}$.
Adding these three equations:
$\sum \tan A \tan B = 3 + \frac{\cos^2 C + \cos^2 A + \cos^2 B}{\cos A \cos B \cos C}$.
Using the identity $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C$ for $\triangle ABC$:
$\sum \tan A \tan B = 3 + \frac{1 - 2 \cos A \cos B \cos C}{\cos A \cos B \cos C} = 3 + \frac{1}{\cos A \cos B \cos C} - 2$.
Substituting $\cos A \cos B \cos C = \frac{1}{5}$:
$\sum \tan A \tan B = 3 + 5 - 2 = 6$.

(A) Given that $\cosh \alpha + \sinh \alpha = e^3$.
Using the definitions $\cosh \alpha = \frac{e^\alpha + e^{-\alpha}}{2}$ and $\sinh \alpha = \frac{e^\alpha - e^{-\alpha}}{2}$,we get:
$\frac{e^\alpha + e^{-\alpha}}{2} + \frac{e^\alpha - e^{-\alpha}}{2} = e^3$
$e^\alpha = e^3 \Rightarrow \alpha = 3$.
Now,substitute $\alpha = 3$ into the expression for $\sinh x$:
$\sinh x = \frac{3}{3+1} = \frac{3}{4}$.
We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + (\frac{3}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
Therefore,$\tanh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{5/4} = \frac{3}{5}$.
Since $\alpha = 3$,we check the options: $\frac{\alpha}{\alpha+2} = \frac{3}{3+2} = \frac{3}{5}$.
Thus,the correct option is $A$.

(A) Given: $\sin \beta=2 \sin \alpha \dots (i)$
and $3 \cos \beta=2 \cos \alpha \Rightarrow \cos \beta=\frac{2}{3} \cos \alpha \dots (ii)$
Squaring and adding the equations:
$\sin^2 \beta + \cos^2 \beta = (2 \sin \alpha)^2 + (\frac{2}{3} \cos \alpha)^2$
$1 = 4 \sin^2 \alpha + \frac{4}{9} \cos^2 \alpha$
$1 = 4 \sin^2 \alpha + \frac{4}{9} (1 - \sin^2 \alpha)$
$1 = 4 \sin^2 \alpha + \frac{4}{9} - \frac{4}{9} \sin^2 \alpha$
$1 - \frac{4}{9} = \frac{32}{9} \sin^2 \alpha$ $\Rightarrow \frac{5}{9} = \frac{32}{9} \sin^2 \alpha$ $\Rightarrow \sin^2 \alpha = \frac{5}{32}$
Similarly,$\cos^2 \alpha = 1 - \frac{5}{32} = \frac{27}{32}$
Now,$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
Substitute $\cos \beta = \frac{2}{3} \cos \alpha$ and $\sin \beta = 2 \sin \alpha$:
$\cos(\alpha + \beta) = \cos \alpha (\frac{2}{3} \cos \alpha) - \sin \alpha (2 \sin \alpha)$
$\cos(\alpha + \beta) = \frac{2}{3} \cos^2 \alpha - 2 \sin^2 \alpha$
$\cos(\alpha + \beta) = \frac{2}{3} (\frac{27}{32}) - 2 (\frac{5}{32}) = \frac{18}{32} - \frac{10}{32} = \frac{8}{32} = \frac{1}{4}$
Therefore,$\sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)} = 4$.

(C) Given: $10 \sin^4 \alpha + 15 \cos^4 \alpha = 6$
Divide by $\cos^4 \alpha$:
$10 \tan^4 \alpha + 15 = 6 \sec^4 \alpha$
$10 \tan^4 \alpha + 15 = 6(1 + \tan^2 \alpha)^2$
$10 \tan^4 \alpha + 15 = 6(1 + 2 \tan^2 \alpha + \tan^4 \alpha)$
$10 \tan^4 \alpha + 15 = 6 + 12 \tan^2 \alpha + 6 \tan^4 \alpha$
$4 \tan^4 \alpha - 12 \tan^2 \alpha + 9 = 0$
$(2 \tan^2 \alpha - 3)^2 = 0$
$\tan^2 \alpha = \frac{3}{2}$
Then,$\cot^2 \alpha = \frac{2}{3}$
Now,$16 \tan^6 \alpha + 27 \cot^6 \alpha = 16(\frac{3}{2})^3 + 27(\frac{2}{3})^3$
$= 16(\frac{27}{8}) + 27(\frac{8}{27}) = 54 + 8 = 62$

(D) Given: $\tan A = \tan \alpha \coth x = \cot \beta \tanh x$ ...$(i)$
From $(i)$,$\tan \alpha = \tan A \tanh x$ and $\tan \beta = \frac{\tanh x}{\tan A}$.
We know $\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.
Substituting the values:
$\tan (\alpha + \beta) = \frac{\tan A \tanh x + \frac{\tanh x}{\tan A}}{1 - (\tan A \tanh x)(\frac{\tanh x}{\tan A})}$
$= \frac{\tanh x (\tan A + \frac{1}{\tan A})}{1 - \tanh^2 x}$
$= \frac{\tanh x (\frac{\tan^2 A + 1}{\tan A})}{\operatorname{sech}^2 x}$
$= \frac{\sinh x}{\cosh x} \cdot \frac{\sec^2 A}{\tan A} \cdot \cosh^2 x$
$= \sinh x \cosh x \cdot \frac{1}{\cos^2 A} \cdot \frac{\cos A}{\sin A}$
$= \frac{2 \sinh x \cosh x}{2 \sin A \cos A} = \frac{\sinh 2x}{\sin 2A} = \sinh 2x \operatorname{cosec} 2A$.

(B) Let $S = \sum_{k=0}^4 \sin^2 \left( (2k+1) \frac{\pi}{20} \right)$.
Expanding the summation,we get:
$S = \sin^2 \frac{\pi}{20} + \sin^2 \frac{3\pi}{20} + \sin^2 \frac{5\pi}{20} + \sin^2 \frac{7\pi}{20} + \sin^2 \frac{9\pi}{20}$.
Using the identity $\sin^2 \theta = \cos^2 \left( \frac{\pi}{2} - \theta \right)$:
$\sin^2 \frac{\pi}{20} = \cos^2 \left( \frac{\pi}{2} - \frac{\pi}{20} \right) = \cos^2 \frac{9\pi}{20}$.
$\sin^2 \frac{3\pi}{20} = \cos^2 \left( \frac{\pi}{2} - \frac{3\pi}{20} \right) = \cos^2 \frac{7\pi}{20}$.
Substituting these into the sum:
$S = \cos^2 \frac{9\pi}{20} + \cos^2 \frac{7\pi}{20} + \sin^2 \frac{5\pi}{20} + \sin^2 \frac{7\pi}{20} + \sin^2 \frac{9\pi}{20}$.
Grouping terms:
$S = (\sin^2 \frac{9\pi}{20} + \cos^2 \frac{9\pi}{20}) + (\sin^2 \frac{7\pi}{20} + \cos^2 \frac{7\pi}{20}) + \sin^2 \frac{\pi}{4}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:
$S = 1 + 1 + \left( \frac{1}{\sqrt{2}} \right)^2 = 2 + \frac{1}{2} = \frac{5}{2}$.

(B) Given $\tan B = \frac{2 \sin A \sin C}{\sin (A+C)}$.
Taking the reciprocal on both sides,we get $\frac{1}{\tan B} = \frac{\sin (A+C)}{2 \sin A \sin C}$.
Using the expansion $\sin (A+C) = \sin A \cos C + \cos A \sin C$,we have $\frac{1}{\tan B} = \frac{\sin A \cos C + \cos A \sin C}{2 \sin A \sin C}$.
This simplifies to $\frac{1}{\tan B} = \frac{1}{2} (\cot A + \cot C) = \frac{1}{2} (\frac{1}{\tan A} + \frac{1}{\tan C})$.
Multiplying by $2$,we get $\frac{2}{\tan B} = \frac{1}{\tan A} + \frac{1}{\tan C}$.
This is the condition for $\tan A, \tan B, \tan C$ to be in Harmonic Progression.

(A) Let the given expression be $E = \frac{1+\cos \theta-\sin \theta}{1+\cos \theta+\sin \theta}+\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$.
Taking the common denominator,we get:
$E = \frac{(1+\cos \theta-\sin \theta)^2 + (1+\cos \theta+\sin \theta)^2}{(1+\cos \theta+\sin \theta)(1+\cos \theta-\sin \theta)}$.
Using the identity $(a-b)^2 + (a+b)^2 = 2(a^2+b^2)$,where $a = 1+\cos \theta$ and $b = \sin \theta$:
Numerator $= 2((1+\cos \theta)^2 + \sin^2 \theta) = 2(1 + 2\cos \theta + \cos^2 \theta + \sin^2 \theta) = 2(1 + 2\cos \theta + 1) = 2(2 + 2\cos \theta) = 4(1+\cos \theta)$.
Denominator $= (1+\cos \theta)^2 - \sin^2 \theta = 1 + 2\cos \theta + \cos^2 \theta - \sin^2 \theta = 1 + 2\cos \theta + \cos^2 \theta - (1 - \cos^2 \theta) = 2\cos^2 \theta + 2\cos \theta = 2\cos \theta(1+\cos \theta)$.
Thus,$E = \frac{4(1+\cos \theta)}{2\cos \theta(1+\cos \theta)} = \frac{2}{\cos \theta} = 2 \sec \theta$.

(A) Given: $m \tan (\theta-30^{\circ})=n \tan (\theta+120^{\circ})$
Since $\tan (\theta+120^{\circ}) = \tan (\theta+30^{\circ}+90^{\circ}) = -\cot (\theta+30^{\circ})$,
$\frac{n}{m} = \frac{\tan (\theta-30^{\circ})}{-\cot (\theta+30^{\circ})} = -\tan (\theta-30^{\circ}) \tan (\theta+30^{\circ})$
Using $\tan (A-B) \tan (A+B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}$,we get:
$\frac{n}{m} = -\left( \frac{\tan^2 \theta - \tan^2 30^{\circ}}{1 - \tan^2 \theta \tan^2 30^{\circ}} \right) = \frac{\tan^2 30^{\circ} - \tan^2 \theta}{1 - \tan^2 \theta \tan^2 30^{\circ}}$
Now,$\frac{m+n}{m-n} = \frac{1 + n/m}{1 - n/m} = \frac{1 + \frac{\tan^2 30^{\circ} - \tan^2 \theta}{1 - \tan^2 \theta \tan^2 30^{\circ}}}{1 - \frac{\tan^2 30^{\circ} - \tan^2 \theta}{1 - \tan^2 \theta \tan^2 30^{\circ}}}$
$= \frac{1 - \tan^2 \theta \tan^2 30^{\circ} + \tan^2 30^{\circ} - \tan^2 \theta}{1 - \tan^2 \theta \tan^2 30^{\circ} - \tan^2 30^{\circ} + \tan^2 \theta} = \frac{(1 - \tan^2 \theta)(1 + \tan^2 30^{\circ})}{(1 + \tan^2 \theta)(1 - \tan^2 30^{\circ})}$
$= \frac{\cos 2\theta}{\cos 2(30^{\circ})} = \frac{\cos 2\theta}{1/2} = 2 \cos 2\theta$

(A) Given $\cos \alpha + \cos \beta = \frac{1}{3} \dots (i)$ and $\sin \alpha + \sin \beta = \frac{1}{4} \dots (ii)$.
Squaring and adding $(i)$ and $(ii)$:
$(\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = (\frac{1}{3})^2 + (\frac{1}{4})^2$
$(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{1}{9} + \frac{1}{16}$
$1 + 1 + 2 \cos (\alpha - \beta) = \frac{16 + 9}{144} = \frac{25}{144}$
$2 + 2 \cos (\alpha - \beta) = \frac{25}{144} \implies 2 \cos (\alpha - \beta) = \frac{25}{144} - 2 = -\frac{263}{144}$.
Alternatively,using sum-to-product formulas:
$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{3} \dots (iii)$
$2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{4} \dots (iv)$
Dividing $(iv)$ by $(iii)$ gives $\tan \frac{\alpha+\beta}{2} = \frac{3}{4}$.
Then $\cos (\alpha+\beta) = \frac{1 - \tan^2 \frac{\alpha+\beta}{2}}{1 + \tan^2 \frac{\alpha+\beta}{2}} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.

(A) Let the given expression be $P = \prod_{k=1}^{7} \left(1+\cos \frac{k\pi}{8}\right)$.
Using the identity $1+\cos \theta = 2\cos^2 \frac{\theta}{2}$,we have $1+\cos \frac{k\pi}{8} = 2\cos^2 \frac{k\pi}{16}$.
Alternatively,note that $\cos(\pi - \theta) = -\cos \theta$,so $1+\cos(\pi - \theta) = 1-\cos \theta$.
Pairing terms: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{7\pi}{8}\right) = (1+\cos \frac{\pi}{8})(1-\cos \frac{\pi}{8}) = 1-\cos^2 \frac{\pi}{8} = \sin^2 \frac{\pi}{8}$.
Similarly,$\left(1+\cos \frac{2\pi}{8}\right)\left(1+\cos \frac{6\pi}{8}\right) = (1+\cos \frac{\pi}{4})(1-\cos \frac{\pi}{4}) = 1-\cos^2 \frac{\pi}{4} = \sin^2 \frac{\pi}{4} = \frac{1}{2}$.
And $\left(1+\cos \frac{3\pi}{8}\right)\left(1+\cos \frac{5\pi}{8}\right) = (1+\cos \frac{3\pi}{8})(1-\cos \frac{3\pi}{8}) = 1-\cos^2 \frac{3\pi}{8} = \sin^2 \frac{3\pi}{8}$.
The middle term is $\left(1+\cos \frac{4\pi}{8}\right) = 1+\cos \frac{\pi}{2} = 1+0 = 1$.
Thus,$P = \sin^2 \frac{\pi}{8} \cdot \sin^2 \frac{3\pi}{8} \cdot \frac{1}{2} \cdot 1$.
Since $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$,$P = \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} \cdot \frac{1}{2} = \frac{1}{4} (2 \sin \frac{\pi}{8} \cos \frac{\pi}{8})^2 \cdot \frac{1}{2} = \frac{1}{4} (\sin \frac{\pi}{4})^2 \cdot \frac{1}{2} = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{16}$.

(B) Given $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.
Substitute $\cos^2 x = 1 - \sin^2 x$:
$3 \sin^4 x + 2(1 - \sin^2 x)^2 = \frac{6}{5}$
$3 \sin^4 x + 2(1 + \sin^4 x - 2 \sin^2 x) = \frac{6}{5}$
$5 \sin^4 x - 4 \sin^2 x + 2 = \frac{6}{5}$
$25 \sin^4 x - 20 \sin^2 x + 10 = 6$
$25 \sin^4 x - 20 \sin^2 x + 4 = 0$
$(5 \sin^2 x - 2)^2 = 0$
$\sin^2 x = \frac{2}{5}$,so $\cos^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.
Now,$\sin 2x = 2 \sin x \cos x$,so $\sin^2 2x = 4 \sin^2 x \cos^2 x = 4 \times \frac{2}{5} \times \frac{3}{5} = \frac{24}{25}$.
$\sin 2x = \frac{2 \sqrt{6}}{5}$ (since $x$ is acute,$2x$ is in the first or second quadrant).
$\cos 2x = 2 \cos^2 x - 1 = 2(\frac{3}{5}) - 1 = \frac{1}{5}$.
$\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{2 \sqrt{6} / 5}{1 / 5} = 2 \sqrt{6}$.

(D) Given $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$.
We need to evaluate $f_1(x) + f_2(x) - f_3(x)$.
$f_1(x) = \frac{1}{2} [\sin^2 x + \cos^2 x] = \frac{1}{2}(1) = \frac{1}{2}$.
$f_2(x) = \frac{1}{4} [\sin^4 x + \cos^4 x] = \frac{1}{4} [(\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x] = \frac{1}{4} [1 - 2 \sin^2 x \cos^2 x] = \frac{1}{4} - \frac{1}{2} \sin^2 x \cos^2 x$.
$f_3(x) = \frac{1}{6} [\sin^6 x + \cos^6 x] = \frac{1}{6} [(\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x)] = \frac{1}{6} [1 - 3 \sin^2 x \cos^2 x]$.
Now,$f_1(x) + f_2(x) - f_3(x) = \frac{1}{2} + (\frac{1}{4} - \frac{1}{2} \sin^2 x \cos^2 x) - (\frac{1}{6} - \frac{1}{2} \sin^2 x \cos^2 x)$.
$= \frac{1}{2} + \frac{1}{4} - \frac{1}{6} - \frac{1}{2} \sin^2 x \cos^2 x + \frac{1}{2} \sin^2 x \cos^2 x$.
$= \frac{6 + 3 - 2}{12} = \frac{7}{12}$.

(B) The given curve is: $x=1+\frac{1}{y^2}$ ...$(i)$
Differentiating with respect to $x$: $1 = -\frac{2}{y^3} \cdot \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -\frac{y^3}{2}$.
At point $A(2,1)$,the slope $m_1 = \left(\frac{dy}{dx}\right)_{(2,1)} = -\frac{1^3}{2} = -\frac{1}{2}$.
The equation of the tangent at $A(2,1)$ is: $(y-1) = -\frac{1}{2}(x-2) \Rightarrow 2y - 2 = -x + 2 \Rightarrow x + 2y = 4$ ...(ii)
To find point $B$,substitute $x = 4-2y$ into the curve equation $(i)$:
$4-2y = 1 + \frac{1}{y^2} \Rightarrow 3-2y = \frac{1}{y^2} \Rightarrow 3y^2 - 2y^3 = 1 \Rightarrow 2y^3 - 3y^2 + 1 = 0$.
Since $A(2,1)$ is on the curve,$y=1$ is a root. Factoring: $(y-1)^2(2y+1) = 0$.
The roots are $y=1$ (at $A$) and $y=-\frac{1}{2}$ (at $B$).
For $y = -\frac{1}{2}$,$x = 4 - 2(-\frac{1}{2}) = 5$. So,$B = (5, -\frac{1}{2})$.
The slope of the tangent at $B(5, -\frac{1}{2})$ is $m_2 = -\frac{(-1/2)^3}{2} = -\frac{-1/8}{2} = \frac{1}{16}$.
The angle $\theta$ between the tangents is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-1/2 - 1/16}{1 + (-1/2)(1/16)} \right| = \left| \frac{-9/16}{1 - 1/32} \right| = \left| \frac{-9/16}{31/32} \right| = \left| -\frac{9}{16} \cdot \frac{32}{31} \right| = \frac{18}{31}$.
Since $\tan \theta = \frac{18}{31} \neq 0$ and $\tan \theta \neq \infty$,the angle is neither $0$ nor $\frac{\pi}{2}$.

(C) Given curve is $y^2 = x + \sin x$ ...$(i)$
By differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1 + \cos x$,which implies $\frac{dy}{dx} = \frac{1 + \cos x}{2y}$.
The slope of the normal is $m = -\frac{1}{dy/dx} = -\frac{2y}{1 + \cos x}$.
For the normal to be parallel to the $Y$-axis,the slope must be undefined,i.e.,$1 + \cos x = 0$.
This implies $\cos x = -1$,which means $x = (2n+1)\pi$ for some integer $n$.
At these points,$\sin x = 0$.
Substituting $\sin x = 0$ into the original equation $(i)$,we get $y^2 = x + 0$,or $y^2 = x$.
This is the equation of a parabola.

(A) Given,the slope of the tangent is $\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$.
At the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,the slope of the tangent $m_t$ is:
$m_t = \frac{(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2}{2(\frac{1}{2})(\frac{\sqrt{3}}{2})} = \frac{\frac{3}{4} - \frac{1}{4}}{\frac{\sqrt{3}}{2}} = \frac{\frac{2}{4}}{\frac{\sqrt{3}}{2}} = \frac{1}{2} \times \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\sqrt{3}$.
The equation of the normal at $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ is:
$y - \frac{\sqrt{3}}{2} = -\sqrt{3}(x - \frac{1}{2})$
$y - \frac{\sqrt{3}}{2} = -\sqrt{3}x + \frac{\sqrt{3}}{2}$
$\sqrt{3}x + y = \sqrt{3}$.

(C) Given the curve $y = \tan^{-1}(\sin \sqrt{x})$.
To find the points where the tangent is parallel to the $x$-axis,we set the derivative $\frac{dy}{dx} = 0$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1 + (\sin \sqrt{x})^2} \cdot \cos \sqrt{x} \cdot \frac{1}{2\sqrt{x}}$.
Setting $\frac{dy}{dx} = 0$ implies $\cos \sqrt{x} = 0$ (since $x > 0$ for the derivative to be defined).
Thus,$\sqrt{x} = (2n+1)\frac{\pi}{2}$ for $n = 0, 1, 2, \dots$.
Given $0 \leq x \leq 8\pi^2$,we have $0 \leq \sqrt{x} \leq 2\sqrt{2}\pi \approx 8.88$.
Possible values for $\sqrt{x}$ are $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$.
At these points,$\sin \sqrt{x} = \sin((2n+1)\frac{\pi}{2}) = \pm 1$.
Therefore,$y = \tan^{-1}(\pm 1) = \pm \frac{\pi}{4}$.

(B) Given the curve $x y=1$,we have $y = \frac{1}{x}$.
Calculating the derivative: $\frac{d y}{d x} = -\frac{1}{x^2}$.
The slope of the tangent at any point $(x, y)$ is $m_t = -\frac{1}{x^2}$.
The slope of the normal $m_n$ is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{m_t} = x^2$.
Since $x^2 \geq 0$ for all real $x$,the slope of the normal must be non-negative $(m_n \geq 0)$.
The given equation of the normal is $(a^2-1) x+a y+(3-a)=0$,which can be rewritten as $a y = -(a^2-1) x - (3-a)$,or $y = -\frac{a^2-1}{a} x - \frac{3-a}{a}$.
The slope of this line is $m = -\frac{a^2-1}{a} = \frac{1-a^2}{a}$.
Since $m \geq 0$,we have $\frac{1-a^2}{a} \geq 0$.
Multiplying by $-1$ reverses the inequality: $\frac{a^2-1}{a} \leq 0$,which is $\frac{(a-1)(a+1)}{a} \leq 0$.
Using the sign scheme (Wavy Curve Method),the values of $a$ satisfying this are $a \in (-\infty, -1] \cup (0, 1]$.

(D) Given the parametric equations of the curve: $x=e^t \cos t$ and $y=e^t \sin t$.
First,we find the derivatives with respect to $t$:
$\frac{dx}{dt} = e^t(\cos t - \sin t)$ and $\frac{dy}{dt} = e^t(\sin t + \cos t)$.
The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin t + \cos t}{\cos t - \sin t}$.
At the point $(1,0)$,we have $y=0$,which implies $e^t \sin t = 0$. Since $e^t \neq 0$,we have $\sin t = 0$,so $t=0$.
At $t=0$,the slope of the tangent is $\frac{dy}{dx} = \frac{\sin 0 + \cos 0}{\cos 0 - \sin 0} = \frac{0+1}{1-0} = 1$.
The slope of the normal $m$ is given by $m = -\frac{1}{dy/dx} = -\frac{1}{1} = -1$.
Since the slope of the normal is $\tan \theta = -1$,and $\theta$ is the angle with the $X$-axis,we have $\theta = \frac{3\pi}{4}$.

(A) Given the curve $y = \sin x$. The derivative is $\frac{dy}{dx} = \cos x$.
The slope of the normal at point $P(h, k)$ is $m = -\frac{1}{\cos h}$.
The equation of the normal passing through the origin $(0, 0)$ is $y - 0 = m(x - 0)$,which simplifies to $y = -\frac{1}{\cos h} x$.
Since the point $P(h, k)$ lies on the normal,we have $k = -\frac{h}{\cos h}$,which implies $\cos h = -\frac{h}{k}$.
Also,since $P(h, k)$ lies on the curve $y = \sin x$,we have $k = \sin h$.
Using the identity $\sin^2 h + \cos^2 h = 1$,we substitute the expressions: $k^2 + (-\frac{h}{k})^2 = 1$.
This simplifies to $k^2 + \frac{h^2}{k^2} = 1$,or $k^4 + h^2 = k^2$.
Replacing $(h, k)$ with $(x, y)$,the locus of $P$ is $x^2 + y^4 = y^2$,which is $x^2 = y^2 - y^4$.

(C) Given the curve $y=x^3$.
The slope of the tangent at point $(\alpha, \beta)$ is $\frac{dy}{dx} = 3x^2$. At $x=\alpha$,the slope is $3\alpha^2$.
The equation of the tangent line at $(\alpha, \beta)$ is $(y-\beta) = 3\alpha^2(x-\alpha)$.
Since $(\alpha_1, \beta_1)$ lies on the curve,$\beta_1 = \alpha_1^3$ and $\beta = \alpha^3$.
Substituting these into the tangent equation: $\alpha_1^3 - \alpha^3 = 3\alpha^2(\alpha_1 - \alpha)$.
Dividing by $(\alpha_1 - \alpha)$ (assuming $\alpha_1 \neq \alpha$): $\alpha_1^2 + \alpha_1\alpha + \alpha^2 = 3\alpha^2$.
$\alpha_1^2 + \alpha_1\alpha - 2\alpha^2 = 0$.
Factoring the quadratic: $(\alpha_1 - \alpha)(\alpha_1 + 2\alpha) = 0$.
Since $\alpha_1 \neq \alpha$,we have $\alpha_1 = -2\alpha$.
Therefore,$\frac{\beta_1}{\beta} = \frac{\alpha_1^3}{\alpha^3} = \left(\frac{\alpha_1}{\alpha}\right)^3 = (-2)^3 = -8$.

(C) Given the curve equation: $x^3 y^2 + \frac{x^2}{y} = 5$.
Differentiating both sides with respect to $x$:
$3x^2 y^2 + 2x^3 y \frac{dy}{dx} + \frac{2x}{y} - \frac{x^2}{y^2} \frac{dy}{dx} = 0$.
Rearranging to solve for $\frac{dy}{dx}$:
$\left(2x^3 y - \frac{x^2}{y^2}\right) \frac{dy}{dx} = -\left(3x^2 y^2 + \frac{2x}{y}\right)$.
$\frac{dy}{dx} = -\frac{3x^2 y^2 + \frac{2x}{y}}{2x^3 y - \frac{x^2}{y^2}} = -\frac{3x^2 y^3 + 2x}{y(2x^3 y^2 - x^2)} = -\frac{x(3xy^3 + 2)}{y(2x^3 y^2 - x^2)}$.
For the tangent to be parallel to the $X$-axis,$\frac{dy}{dx} = 0$,which implies $3xy^3 + 2 = 0$.
This is the equation $f(x, y) = 0$.
Testing the given options:
For $\left(-2, \frac{1}{\sqrt[3]{3}}\right)$:
$3(-2)\left(\frac{1}{\sqrt[3]{3}}\right)^3 + 2 = 3(-2)\left(\frac{1}{3}\right) + 2 = -2 + 2 = 0$.
Thus,the point $\left(-2, \frac{1}{\sqrt[3]{3}}\right)$ satisfies the equation.

(C) Given the curve $x=e^{\sin y}$. Taking the natural logarithm on both sides,we get $\log x = \sin y$.
Differentiating both sides with respect to $x$,we have $\frac{1}{x} = \cos y \frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{x \cos y}$.
The slope of the tangent at $(1,0)$ is $\left(\frac{dy}{dx}\right)_{(1,0)} = \frac{1}{1 \cdot \cos 0} = \frac{1}{1 \cdot 1} = 1$.
The slope of the normal at $(1,0)$ is $m = -\frac{1}{\text{slope of tangent}} = -\frac{1}{1} = -1$.
The equation of the normal at $(1,0)$ is $y - 0 = -1(x - 1)$,which simplifies to $y = -x + 1$ or $x + y = 1$.
This line intersects the $x$-axis at $A(1,0)$ and the $y$-axis at $B(0,1)$.
The triangle formed by the normal with the coordinate axes is $\triangle OAB$,where $O$ is the origin $(0,0)$.
The area of $\triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \text{ sq units}$.

(B) Given the curve equation: $(x^2+1)(y-3)=x$ ... $(i)$
Differentiating both sides with respect to $x$:
$(x^2+1) \frac{dy}{dx} + (2x)(y-3) = 1$
$\frac{dy}{dx} = \frac{1 - 2x(y-3)}{x^2+1}$
Since the tangent is a horizontal line,the slope $\frac{dy}{dx} = 0$.
Therefore,$1 - 2x(y-3) = 0 \Rightarrow 2x(y-3) = 1$ ... $(ii)$
From $(i)$,we have $(y-3) = \frac{x}{x^2+1}$.
Substituting this into $(ii)$: $2x \left( \frac{x}{x^2+1} \right) = 1$
$2x^2 = x^2 + 1 \Rightarrow x^2 = 1$.
Since $P$ lies in the first quadrant,$x = 1$.
Substituting $x=1$ into $(i)$: $(1^2+1)(y-3) = 1 \Rightarrow 2(y-3) = 1 \Rightarrow y-3 = \frac{1}{2} \Rightarrow y = \frac{7}{2}$.
The point $P$ is $(1, \frac{7}{2})$.
Since the tangent is horizontal,the normal is a vertical line passing through $x=1$.
The equation of the normal is $x = 1$.

(C) The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = xy$.
At the point $(\sqrt{2}, e)$,the slope of the tangent is $m = \sqrt{2}e$.
The slope of the normal at this point is $m' = -\frac{1}{m} = -\frac{1}{\sqrt{2}e}$.
The equation of the normal at $(\sqrt{2}, e)$ is given by $(y - e) = m'(x - \sqrt{2})$.
Substituting $m'$,we get $y - e = -\frac{1}{\sqrt{2}e}(x - \sqrt{2})$.
Multiplying by $\sqrt{2}e$,we get $\sqrt{2}ey - \sqrt{2}e^2 = -x + \sqrt{2}$.
Rearranging the terms,$x + \sqrt{2}ey = \sqrt{2} + \sqrt{2}e^2$.
Dividing by $\sqrt{2}(1 + e^2)$,we get $\frac{x}{\sqrt{2}(1 + e^2)} + \frac{\sqrt{2}ey}{\sqrt{2}(1 + e^2)} = 1$.
Comparing this with $ax + by = 1$,we have $a = \frac{1}{\sqrt{2}(1 + e^2)}$ and $b = \frac{\sqrt{2}e}{\sqrt{2}(1 + e^2)} = \frac{e}{1 + e^2}$.
Therefore,$\frac{b}{a} = \frac{e}{1 + e^2} \times \sqrt{2}(1 + e^2) = \sqrt{2}e$.

(C) Given the curve $y = \sin x$. The slope of the tangent at any point $(x_1, y_1)$ is given by $\frac{dy}{dx} = \cos x_1$.
The equation of the tangent line at $(x_1, y_1)$ is $(y - y_1) = \cos x_1(x - x_1)$.
Since this tangent passes through the point $(0, \pi)$,we substitute these coordinates into the equation:
$(\pi - y_1) = \cos x_1(0 - x_1) = -x_1 \cos x_1$.
Since $y_1 = \sin x_1$,we have $\cos x_1 = \pm \sqrt{1 - \sin^2 x_1} = \pm \sqrt{1 - y_1^2}$.
Substituting this into the equation: $(\pi - y_1) = -x_1(\pm \sqrt{1 - y_1^2})$.
Squaring both sides,we get $(\pi - y_1)^2 = x_1^2(1 - y_1^2)$.
Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $x^2(1 - y^2) = (y - \pi)^2$.

(C) Given the curve $y = \cosh x$.
To find the point nearest to the origin $(0, 0)$,we minimize the squared distance $D^2 = x^2 + y^2 = x^2 + (\cosh x)^2$.
Let $f(x) = x^2 + \cosh^2 x$.
Taking the derivative,$f'(x) = 2x + 2 \cosh x \sinh x = 2x + \sinh(2x)$.
Setting $f'(x) = 0$,we get $2x + \sinh(2x) = 0$.
The only solution to this equation is $x = 0$.
At $x = 0$,$y = \cosh(0) = 1$.
The slope of the tangent at $(0, 1)$ is $y' = \sinh(0) = 0$.
The slope of the normal $m$ is given by $m = -\frac{1}{y'} = -\frac{1}{0}$,which is undefined (vertical line).
The equation of a vertical line passing through $(0, 1)$ is $x = 0$.

(C) Given the family of curves $y = x^n \log x$.
For the tangent $y = x - 1$ to be common to all curves at a fixed point $(\alpha, \beta)$,the slope of the tangent must be $1$ (since the slope of $y = x - 1$ is $1$).
Calculating the derivative: $\frac{dy}{dx} = n x^{n-1} \log x + x^n \cdot \frac{1}{x} = x^{n-1} (n \log x + 1)$.
At the point of tangency $(\alpha, \beta)$,the slope is $1$:
$\alpha^{n-1} (n \log \alpha + 1) = 1$.
For this to hold for all $n$,we test the condition at $\alpha = 1$:
$1^{n-1} (n \log 1 + 1) = 1 \cdot (0 + 1) = 1$.
This is independent of $n$.
When $\alpha = 1$,the value of $y$ on the curve is $\beta = 1^n \log 1 = 0$.
Thus,the fixed point is $(1, 0)$.
Therefore,$\alpha + \beta = 1 + 0 = 1$.

(D) Given that $px+my+n=0$ is the tangent to the curve $y=f(x)$ at $x=\alpha$.
The slope of the tangent line $px+my+n=0$ is $m_t = -\frac{p}{m}$.
Therefore,$f'(\alpha) = -\frac{p}{m}$.
Now,we need to find $\frac{d}{dx} [f(\alpha e^{2x})]$ at $x=0$.
Using the chain rule:
$\frac{d}{dx} [f(\alpha e^{2x})] = f'(\alpha e^{2x}) \cdot \frac{d}{dx}(\alpha e^{2x}) = f'(\alpha e^{2x}) \cdot (2\alpha e^{2x})$.
At $x=0$,the expression becomes:
$f'(\alpha e^0) \cdot (2\alpha e^0) = f'(\alpha) \cdot (2\alpha)$.
Substituting $f'(\alpha) = -\frac{p}{m}$:
$(-\frac{p}{m}) \cdot (2\alpha) = -\frac{2p\alpha}{m}$.

(D) Given that $|f(x) - f(y)| \leq 2|x - y|^{\frac{3}{2}}$.
Dividing both sides by $|x - y|$ (where $x \neq y$),we get $\left|\frac{f(x) - f(y)}{x - y}\right| \leq 2|x - y|^{\frac{1}{2}}$.
Taking the limit as $x \rightarrow y$,the left side becomes the definition of the derivative $|f'(y)|$.
Thus,$|f'(y)| \leq 2 \lim_{x \rightarrow y} |x - y|^{\frac{1}{2}} = 0$.
Since the absolute value cannot be negative,we must have $|f'(y)| = 0$,which implies $f'(y) = 0$ for all $y \in R$.
This means $f(x)$ is a constant function,$f(x) = c$.
Given $f(0) = 1$,we have $c = 1$,so $f(x) = 1$.
Therefore,$\int_0^1 f^2(x) dx = \int_0^1 (1)^2 dx = \int_0^1 1 dx = [x]_0^1 = 1 - 0 = 1$.

(C) Let $f(x) = x^4+ax^3+\frac{3x^2}{2}+1$.
First,find the first derivative: $f'(x) = 4x^3+3ax^2+3x$.
Next,find the second derivative: $f''(x) = 12x^2+6ax+3$.
We are given that $f''(x) > 0$ for all real $x$,so $12x^2+6ax+3 > 0$.
Dividing by $3$,we get $4x^2+2ax+1 > 0$.
For a quadratic $Ax^2+Bx+C > 0$ to hold for all $x$,the discriminant $D$ must be less than $0$ and $A > 0$.
Here $A = 4 > 0$,so we require $D < 0$.
$D = (2a)^2 - 4(4)(1) < 0$.
$4a^2 - 16 < 0$.
$a^2 < 4$.
This implies $-2 < a < 2$.
The maximum value of $a$ is $2$.

(D) To determine which function is decreasing in the interval $\left(\frac{1}{e}, e\right)$,we check the sign of the derivative $f'(x)$ for each function.
For option $(A)$: $f(x) = \frac{\log x}{x}$,$f'(x) = \frac{1 - \log x}{x^2}$. For $x \in \left(\frac{1}{e}, e\right)$,$\log x$ ranges from $-1$ to $1$. Thus,$f'(x) > 0$ for $x < e$,so it is increasing.
For option $(B)$: $f(x) = x^2 \log x$,$f'(x) = 2x \log x + x = x(2 \log x + 1)$. For $x > \frac{1}{e}$,$f'(x) > 0$,so it is increasing.
For option $(C)$: $f(x) = x \log x$,$f'(x) = \log x + 1$. For $x > \frac{1}{e}$,$\log x > -1$,so $f'(x) > 0$,so it is increasing.
For option $(D)$: $f(x) = x^{-x}$. Let $y = x^{-x}$,then $\log y = -x \log x$. Differentiating with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = -(\log x + x \cdot \frac{1}{x}) = -(\log x + 1)$. Thus,$f'(x) = -x^{-x}(1 + \log x)$. In the interval $\left(\frac{1}{e}, e\right)$,$\log x > -1$,so $(1 + \log x) > 0$. Since $x^{-x} > 0$,$f'(x) < 0$. Therefore,$f(x) = x^{-x}$ is a decreasing function.

(D) Given the curve $y = x^3 - ax^2 + x + 1$.
The slope of the tangent is given by $\frac{dy}{dx} = 3x^2 - 2ax + 1$.
Since the tangent makes an acute angle with the positive direction of the $X$-axis,the slope must be positive for all $x \in R$.
Thus,$3x^2 - 2ax + 1 > 0$ for all $x \in R$.
For a quadratic expression $Ax^2 + Bx + C > 0$ to be true for all $x \in R$,we must have $A > 0$ and the discriminant $D < 0$.
Here,$A = 3 > 0$,which is satisfied.
Now,we require $D = B^2 - 4AC < 0$.
Substituting the values,$(-2a)^2 - 4(3)(1) < 0$.
$4a^2 - 12 < 0$.
$a^2 - 3 < 0$.
$(a - \sqrt{3})(a + \sqrt{3}) < 0$.
Therefore,$a \in (-\sqrt{3}, \sqrt{3})$.

(C) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y=x^2+3x+4$,we have $k=h^2+3h+4$ ... $(i)$.
The slope of the tangent at $(h, k)$ is $\frac{dy}{dx} = 2x+3$. At $(h, k)$,the slope is $m = 2h+3$.
The equation of the tangent at $(h, k)$ is $y-k = (2h+3)(x-h)$.
Since the tangent passes through $(0, 0)$,we substitute $x=0$ and $y=0$:
$0-k = (2h+3)(0-h) \Rightarrow -k = -2h^2-3h \Rightarrow k = 2h^2+3h$ ... $(ii)$.
Equating $(i)$ and $(ii)$:
$h^2+3h+4 = 2h^2+3h$
$h^2 = 4 \Rightarrow h = \pm 2$.
If $h = -2$,$k = (-2)^2+3(-2)+4 = 4-6+4 = 2$. So,$(\alpha, \beta) = (-2, 2)$.
If $h = 2$,$k = (2)^2+3(2)+4 = 4+6+4 = 14$. So,$(\gamma, \delta) = (2, 14)$.
Thus,$\beta+\delta = 2+14 = 16$.

(C) Given $f(x) = \max\{\cos x, \sin x, 0\}$.
In the interval $[0, 2\pi]$,the function $f(x)$ is defined as:
$f(x) = \cos x$ for $x \in [0, \pi/4]$
$f(x) = \sin x$ for $x \in [\pi/4, 5\pi/4]$
$f(x) = 0$ for $x \in [5\pi/4, 2\pi]$
Checking for non-differentiability at the intersection points:
$1$. At $x = \pi/4$,$\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$. The derivatives are $-\sin(\pi/4) = -1/\sqrt{2}$ and $\cos(\pi/4) = 1/\sqrt{2}$. Since they are not equal,$f(x)$ is not differentiable at $x = \pi/4$.
$2$. At $x = 5\pi/4$,$\sin(5\pi/4) = -1/\sqrt{2}$ and $0$. Since $\sin(5\pi/4) < 0$,the function is $0$. The transition from $\sin x$ to $0$ occurs at $x = \pi$ (where $\sin x = 0$). At $x = \pi$,$\sin(\pi) = 0$ and the derivative is $\cos(\pi) = -1$. The derivative of $0$ is $0$. Since $-1 \neq 0$,it is not differentiable at $x = \pi$.
$3$. At $x = 2\pi$,$\cos(2\pi) = 1$ and $0$. The transition from $0$ to $\cos x$ occurs at $x = 2\pi - \pi/2 = 3\pi/2$ is not correct,actually $\cos x = 0$ at $x = 3\pi/2$. The function $f(x)$ is $0$ on $[5\pi/4, 2\pi]$ and $\cos x$ starts being positive at $x = 3\pi/2$. At $x = 3\pi/2$,$\cos(3\pi/2) = 0$ and the derivative is $-\sin(3\pi/2) = 1$. Since $1 \neq 0$,it is not differentiable at $x = 3\pi/2$.
Thus,there are $3$ points of non-differentiability in every interval of length $2\pi$.
In $(0, 2024\pi)$,there are $1012$ such intervals.
Total points of non-differentiability $= 3 \times 1012 = 3036$.
Given $1012k = 3036$,we get $k = 3$.

(C) Given $f(x) = px^3 + qx^2 + rx + t$.
The derivative is $f'(x) = 3px^2 + 2qx + r$ ... $(i)$
Since $f(x)$ has local extrema at $x = -2$ and $x = 2$,$f'(x) = k(x+2)(x-2) = k(x^2 - 4)$ ... (ii)
Comparing $(i)$ and (ii),we get $3p = k$,$2q = 0$,and $r = -4k$.
Thus,$q = 0$ and $r = -4(3p) = -12p$.
The second derivative is $f''(x) = 6px + 2q = 6px$.
For local minimum at $x = -2$,$f''(-2) = -12p > 0$,which implies $p < 0$.
Given $p$ is a root of $9x^2 - 1 = 0$,so $x = \pm \frac{1}{3}$.
Since $p < 0$,we have $p = -\frac{1}{3}$.
Then $r = -12(-\frac{1}{3}) = 4$.
Therefore,$p + q + r = -\frac{1}{3} + 0 + 4 = \frac{11}{3}$.

(D) Let $l$ and $b$ be the length and width of the rectangle.
Given that the perimeter is $48 \text{ cm}$.
$2(l + b) = 48 \Rightarrow l + b = 24 \Rightarrow b = 24 - l$ ... $(i)$
When the rectangle is rotated about side $b$,the radius of the cylinder is $r = l$ and the height is $h = b$.
The volume of the cylinder is $V = \pi r^2 h = \pi l^2 b$.
Substituting $b$ from $(i)$: $V = \pi l^2(24 - l) = 24\pi l^2 - \pi l^3$.
To find the maximum volume,differentiate with respect to $l$:
$\frac{dV}{dl} = 48\pi l - 3\pi l^2$.
Set $\frac{dV}{dl} = 0$ for critical points:
$3\pi l(16 - l) = 0 \Rightarrow l = 0$ or $l = 16$.
Since $l$ must be positive,$l = 16$.
Check the second derivative: $\frac{d^2V}{dl^2} = 48\pi - 6\pi l$.
At $l = 16$,$\frac{d^2V}{dl^2} = 48\pi - 96\pi = -48\pi < 0$.
Since the second derivative is negative,the volume is maximum at $l = 16$.
Then $b = 24 - 16 = 8$.
Thus,the dimensions of the rectangle are $8 \text{ cm}$ and $16 \text{ cm}$.

(A) Given the function $f(x) = \frac{4}{3}x^3 - 4x$ on the interval $[0, 2]$.
First,find the critical points by setting the derivative to zero:
$f'(x) = 4x^2 - 4 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.
Since we are considering the interval $[0, 2]$,we only consider $x = 1$ as a critical point within the interval.
Now,evaluate the function at the critical point and the endpoints of the interval:
$f(0) = \frac{4}{3}(0)^3 - 4(0) = 0$.
$f(1) = \frac{4}{3}(1)^3 - 4(1) = \frac{4}{3} - 4 = -\frac{8}{3}$.
$f(2) = \frac{4}{3}(2)^3 - 4(2) = \frac{32}{3} - 8 = \frac{32 - 24}{3} = \frac{8}{3}$.
Comparing these values,the global minimum is $-\frac{8}{3}$ and the global maximum is $\frac{8}{3}$.
The sum of the global minimum and global maximum values is $\frac{8}{3} + (-\frac{8}{3}) = 0$.

(C) Given that $A(1,15), B(3,-12), C(6,12)$ are three consecutive turning points of a continuous curve $y=f(x)$ and the curve intersects the $x$-axis at $x=\alpha$ and $x=\beta$.
From the given graph,it is clear that:
$1 < \alpha < 3$ and $3 < \beta < 6$.
We want to find the range for $|\beta-\alpha|$.
Since $1 < \alpha < 3$,we have $-3 < -\alpha < -1$.
Also,$3 < \beta < 6$.
Adding these inequalities,we get:
$3 - 3 < \beta - \alpha < 6 - 1$
$0 < \beta - \alpha < 5$.
Therefore,$|\beta-\alpha| < 5$.

(B) Given $f(x) = x^2 + \frac{54}{x}$.
First,find the derivative: $f'(x) = 2x - \frac{54}{x^2}$.
For $x \in (-\infty, 0)$,$x^2 > 0$ and $x < 0$,so $2x < 0$ and $-\frac{54}{x^2} < 0$. Thus,$f'(x) < 0$,which means $f(x)$ is decreasing in $(-\infty, 0)$.
For critical points,set $f'(x) = 0$: $2x - \frac{54}{x^2} = 0 \Rightarrow 2x^3 = 54 \Rightarrow x^3 = 27 \Rightarrow x = 3$.
Since the only critical point $x = 3$ lies in $(0, \infty)$,there is no critical point in $(-\infty, 0)$.
Therefore,$f(x)$ has neither maximum nor minimum in $(-\infty, 0)$.

(B) Using the $AM$-$GM$ inequality for two positive terms $a^2 x^4$ and $b^2 y^4$:
$\frac{a^2 x^4 + b^2 y^4}{2} \ge \sqrt{(a^2 x^4)(b^2 y^4)}$
Given $a^2 x^4 + b^2 y^4 = c^6$,we substitute:
$\frac{c^6}{2} \ge \sqrt{a^2 b^2 x^4 y^4}$
$\frac{c^6}{2} \ge ab x^2 y^2$
Squaring both sides:
$\frac{c^{12}}{4} \ge a^2 b^2 x^4 y^4$
$x^4 y^4 \le \frac{c^{12}}{4 a^2 b^2}$
Thus,the maximum value is $\frac{c^{12}}{4 a^2 b^2}$.

(B) Given $f(x) = 3x + \frac{12}{x}$.
First,find the derivative: $f'(x) = 3 - \frac{12}{x^2}$.
For critical points,set $f'(x) = 0$:
$3 - \frac{12}{x^2} = 0 \implies x^2 = 4 \implies x = 2, -2$.
Now,find the second derivative: $f''(x) = \frac{24}{x^3}$.
At $x = 2$,$f''(2) = \frac{24}{8} = 3 > 0$ (Local minimum).
At $x = -2$,$f''(-2) = \frac{24}{-8} = -3 < 0$ (Local maximum).
The local maximum value $M = f(-2) = 3(-2) + \frac{12}{-2} = -6 - 6 = -12$.
We need to evaluate $\lim_{x \rightarrow M} f(x) = \lim_{x \rightarrow -12} (3x + \frac{12}{x})$.
Substituting $x = -12$: $3(-12) + \frac{12}{-12} = -36 - 1 = -37$.

(A) Given $f(x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}$.
To find the extreme value,we differentiate $f(x)$ with respect to $x$:
$f'(x) = -\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2}$.
Setting $f'(x) = 0$ for critical points:
$-\frac{4 \cos x}{\sin^2 x} + \frac{\cos x}{(1-\sin x)^2} = 0$.
Since $x \in (0, \pi/2)$,$\cos x \neq 0$,so we can divide by $\cos x$:
$-\frac{4}{\sin^2 x} + \frac{1}{(1-\sin x)^2} = 0$.
$\frac{1}{(1-\sin x)^2} = \frac{4}{\sin^2 x}$.
Taking the square root on both sides:
$\frac{1}{1-\sin x} = \frac{2}{\sin x}$ (since $\sin x > 0$ and $1-\sin x > 0$ in the given interval).
$\sin x = 2 - 2 \sin x$.
$3 \sin x = 2 \Rightarrow \sin x = \frac{2}{3}$.
Now,substitute $\sin x = \frac{2}{3}$ into $f(x)$:
$f(x) = \frac{4}{2/3} + \frac{1}{1-2/3} = 4 \times \frac{3}{2} + \frac{1}{1/3} = 6 + 3 = 9$.

(A) Let the height of the cone be $h$ and the radius of its base be $r$. The sphere has radius $R$. From the geometry of the cone inscribed in a sphere,we have the relation $r^2 = R^2 - (h - R)^2 = 2hR - h^2$.
The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{1}{3} \pi (2Rh^2 - h^3)$.
To find the maximum volume,we differentiate $V$ with respect to $h$ and set it to zero:
$\frac{dV}{dh} = \frac{1}{3} \pi (4Rh - 3h^2) = 0$.
Since $h \neq 0$,we have $4R - 3h = 0$,which gives $h = \frac{4R}{3}$.
Thus,$k = \frac{4}{3}$.
The volume of the cone at $h = \frac{4R}{3}$ is $V_{cone} = \frac{1}{3} \pi (2R(\frac{4R}{3})^2 - (\frac{4R}{3})^3) = \frac{1}{3} \pi (\frac{32R^3}{9} - \frac{64R^3}{27}) = \frac{1}{3} \pi (\frac{96R^3 - 64R^3}{27}) = \frac{32}{81} \pi R^3$.
The volume of the sphere is $V_{sphere} = \frac{4}{3} \pi R^3$.
The ratio of the volume of the cone to the volume of the sphere is $\frac{\frac{32}{81} \pi R^3}{\frac{4}{3} \pi R^3} = \frac{32}{81} \times \frac{3}{4} = \frac{8}{27}$.

(C) Given $f(x) = \int \frac{dx}{x^2+(\sqrt{2})^2}$.
Using the standard integral formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$f(x) = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{x}{\sqrt{2}}) + C$.
Given $f(\sqrt{2}) = 0$,substitute $x = \sqrt{2}$:
$0 = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{\sqrt{2}}{\sqrt{2}}) + C$
$0 = \frac{1}{\sqrt{2}} \tan^{-1}(1) + C$
$0 = \frac{1}{\sqrt{2}} (\frac{\pi}{4}) + C$
$C = -\frac{\pi}{4\sqrt{2}}$.
Now,find $f(0)$:
$f(0) = \frac{1}{\sqrt{2}} \tan^{-1}(\frac{0}{\sqrt{2}}) - \frac{\pi}{4\sqrt{2}}$
$f(0) = \frac{1}{\sqrt{2}} (0) - \frac{\pi}{4\sqrt{2}} = -\frac{\pi}{4\sqrt{2}}$.

(C) Given $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 2x^7 + 1)^2} dx$.
Divide the numerator and denominator by $x^{14}$:
$f(x) = \int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + 2 + x^{-7})^2} dx$.
Let $u = x^{-5} + x^{-7} + 2$.
Then $du = (-5x^{-6} - 7x^{-8}) dx$,which implies $-(5x^{-6} + 7x^{-8}) dx = du$.
Substituting this into the integral:
$f(x) = -\int \frac{1}{u^2} du = \frac{1}{u} + C = \frac{1}{x^{-5} + x^{-7} + 2} + C$.
Since $f(0) = 0$,we evaluate the limit as $x \to 0^+$.
As $x \to 0^+$,$x^{-5} + x^{-7} \to \infty$,so $\frac{1}{x^{-5} + x^{-7} + 2} \to 0$.
Thus,$0 = 0 + C$,which gives $C = 0$.
Therefore,$f(x) = \frac{1}{x^{-5} + x^{-7} + 2} = \frac{x^7}{1 + x^2 + 2x^7}$.
Evaluating at $x = 1$:
$f(1) = \frac{1^7}{1 + 1^2 + 2(1)^7} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$.

(B) We are given the integral $I = \int 3^{-\log _9 x^2} d x$.
Using the property of logarithms $n \log_b a = \log_b a^n$,we have $-\log _9 x^2 = \log _9 (x^2)^{-1} = \log _9 (\frac{1}{x^2})$.
Substituting this into the integral,we get $I = \int 3^{\log _9 (\frac{1}{x^2})} d x$.
Using the property $a^{\log _b c} = c^{\log _b a}$,we rewrite $3^{\log _9 (\frac{1}{x^2})}$ as $(\frac{1}{x^2})^{\log _9 3}$.
Since $\log _9 3 = \log _{3^2} 3 = \frac{1}{2} \log _3 3 = \frac{1}{2}$,the expression becomes $(\frac{1}{x^2})^{\frac{1}{2}} = \frac{1}{x}$.
Thus,$I = \int \frac{1}{x} d x = \log |x| + C$.

(A) Let $I = \int \frac{3x+2}{4x^2+4x+5} dx$.
We express the numerator as a derivative of the denominator:
$3x+2 = \frac{3}{8}(8x+4) + (2 - \frac{12}{8}) = \frac{3}{8}(8x+4) + \frac{1}{2}$.
Substituting this into the integral:
$I = \int \frac{\frac{3}{8}(8x+4) + \frac{1}{2}}{4x^2+4x+5} dx = \frac{3}{8} \int \frac{8x+4}{4x^2+4x+5} dx + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$.
For the first part,let $u = 4x^2+4x+5$,then $du = (8x+4)dx$.
$I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \int \frac{1}{(2x+1)^2 + 2^2} dx$.
Using the formula $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$:
$I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \tan^{-1}(\frac{2x+1}{2}) + C$.
$I = \frac{3}{8} \log(4x^2+4x+5) + \frac{1}{8} \tan^{-1}(x+\frac{1}{2}) + C$.
Comparing with the given form,$A = \frac{3}{8}$ and $B = \frac{1}{8}$.

(C) We know the standard integral formula: $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Here,let $f(x) = \log x$.
Then,$f'(x) = \frac{1}{x}$.
Substituting these into the formula:
$\int e^x \left( \log x + \frac{1}{x} \right) dx = e^x \log x + C$.
However,looking at the provided options,the question likely intended to be $\int e^x \left( \log x + \frac{1}{x} \right) dx$. If the question is $\int e^x \left( \log x + \frac{1}{x} \right) dx$,the result is $e^x \log x + C$. If the question is $\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx$,the result is $e^x \frac{1}{x} + C$. Given the structure of the options,there is a mismatch. Assuming the standard form $\int e^x (f(x) + f'(x)) dx$,the closest match for the form $e^x(\log x + \dots)$ is option $C$ if the derivative was different.

(A) Given $\int f(x) dx = \psi(x)$.
Let $I = \int x^5 f(x^3) dx = \int x^3 f(x^3) x^2 dx$.
Substitute $t = x^3$,then $dt = 3x^2 dx$,which implies $x^2 dx = \frac{dt}{3}$.
Substituting these into the integral,we get $I = \int t f(t) \frac{dt}{3} = \frac{1}{3} \int t f(t) dt$.
Using integration by parts,where $u = t$ and $dv = f(t) dt$ (so $du = dt$ and $v = \psi(t)$):
$I = \frac{1}{3} [t \psi(t) - \int \psi(t) dt]$.
Now,substitute $t = x^3$ back into the expression:
$I = \frac{1}{3} [x^3 \psi(x^3) - \int \psi(x^3) (3x^2 dx)]$.
$I = \frac{1}{3} x^3 \psi(x^3) - \int x^2 \psi(x^3) dx$.

(B) Given $f(x)=1+2^x \log 2$.
Since $g(x)$ is an antiderivative of $f(x)$,we have $g(x) = \int f(x) dx$.
$g(x) = \int (1+2^x \log 2) dx = x + \frac{2^x \log 2}{\log 2} + c = x + 2^x + c$.
Thus,$y = x + 2^x + c$ ... $(i)$.
The curve passes through $\left(-1, \frac{1}{2}\right)$,so $\frac{1}{2} = -1 + 2^{-1} + c$.
$\frac{1}{2} = -1 + \frac{1}{2} + c \implies c = 1$.
Substituting $c=1$ in $(i)$,we get $y = x + 2^x + 1$.
To find the intersection with the $Y$-axis,set $x=0$:
$y = 0 + 2^0 + 1 = 1 + 1 = 2$.
Therefore,the curve meets the $Y$-axis at $(0,2)$.

(A) Given equation: $5 f(x)+3 f\left(\frac{1}{x}\right)=2-\frac{1}{x}$ ...$(i)$
Replace $x$ with $\frac{1}{x}$ in equation $(i)$:
$5 f\left(\frac{1}{x}\right)+3 f(x)=2-x$ ...(ii)
To eliminate $f(x)$,multiply equation (ii) by $5$ and equation $(i)$ by $3$:
$25 f\left(\frac{1}{x}\right)+15 f(x)=10-5 x$ ...(iii)
$15 f(x)+9 f\left(\frac{1}{x}\right)=6-\frac{3}{x}$ ...(iv)
Subtract equation (iv) from equation (iii):
$(25-9) f\left(\frac{1}{x}\right) = (10-5 x) - (6-\frac{3}{x})$
$16 f\left(\frac{1}{x}\right) = 4-5 x+\frac{3}{x}$
$f\left(\frac{1}{x}\right) = \frac{1}{16} \left(4-5 x+\frac{3}{x}\right)$
Now,calculate the integral:
$\int_1^2 f\left(\frac{1}{x}\right) d x = \int_1^2 \frac{1}{16} \left(4-5 x+\frac{3}{x}\right) d x$
$= \frac{1}{16} \left[4 x-\frac{5 x^2}{2}+3 \ln |x|\right]_1^2$
$= \frac{1}{16} \left[ \left(4(2)-\frac{5(2)^2}{2}+3 \ln 2\right) - \left(4(1)-\frac{5(1)^2}{2}+3 \ln 1\right) \right]$
$= \frac{1}{16} \left[ (8-10+3 \ln 2) - (4-2.5+0) \right]$
$= \frac{1}{16} \left[ (-2+3 \ln 2) - 1.5 \right]$
$= \frac{1}{16} \left[ 3 \ln 2 - 3.5 \right] = \frac{1}{16} \left[ 3 \ln 2 - \frac{7}{2} \right]$
$= \frac{6 \ln 2-7}{32}$

(A) We have the integral $I = \int e^{-2x} \left( \frac{1 - \sin 2x}{1 + \cos 2x} \right) dx$.
Using trigonometric identities,$1 - \sin 2x = 1 - 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$.
So,the expression becomes $\frac{1 - 2 \sin x \cos x}{2 \cos^2 x} = \frac{1}{2} \sec^2 x - \tan x$.
Thus,$I = \int e^{-2x} (\frac{1}{2} \sec^2 x - \tan x) dx = \frac{1}{2} \int e^{-2x} \sec^2 x dx - \int e^{-2x} \tan x dx$.
Applying integration by parts to the first term: $\int e^{-2x} \sec^2 x dx = e^{-2x} \tan x - \int (-2e^{-2x}) \tan x dx = e^{-2x} \tan x + 2 \int e^{-2x} \tan x dx$.
Substituting this back into $I$: $I = \frac{1}{2} [e^{-2x} \tan x + 2 \int e^{-2x} \tan x dx] - \int e^{-2x} \tan x dx$.
$I = \frac{1}{2} e^{-2x} \tan x + \int e^{-2x} \tan x dx - \int e^{-2x} \tan x dx = \frac{1}{2} e^{-2x} \tan x + C$.

(C) Let $I = \int \frac{4 e^x + 6 e^{-x}}{9 e^x - 4 e^{-x}} d x = \int \frac{4 e^{2 x} + 6}{9 e^{2 x} - 4} d x$.
We express the numerator as $4 e^{2 x} + 6 = A(9 e^{2 x} - 4) + B \frac{d}{d x}(9 e^{2 x} - 4)$.
$4 e^{2 x} + 6 = A(9 e^{2 x} - 4) + B(18 e^{2 x})$.
Equating the coefficients of $e^{2 x}$ and the constant terms:
$9 A + 18 B = 4$ and $-4 A = 6$.
From $-4 A = 6$,we get $A = -\frac{3}{2}$.
Substituting $A$ into the first equation: $9(-\frac{3}{2}) + 18 B = 4 \Rightarrow -\frac{27}{2} + 18 B = 4 \Rightarrow 18 B = 4 + \frac{27}{2} = \frac{35}{2} \Rightarrow B = \frac{35}{36}$.
Thus,$I = \int \left( A + B \frac{18 e^{2 x}}{9 e^{2 x} - 4} \right) d x = A x + B \log |9 e^{2 x} - 4| + C$.
Therefore,$(A, B) = (-\frac{3}{2}, \frac{35}{36})$.

(A) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}} dx}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} \dots (1)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}} dx}{(\sin(\frac{\pi}{2}-x))^{\sqrt{2}}+(\cos(\frac{\pi}{2}-x))^{\sqrt{2}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\cos x)^{\sqrt{2}} dx}{(\cos x)^{\sqrt{2}}+(\sin x)^{\sqrt{2}}} \dots (2)$
Adding $(1)$ and $(2)$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}}{(\sin x)^{\sqrt{2}}+(\cos x)^{\sqrt{2}}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$I = \frac{\pi}{12}$. Thus,Assertion $(A)$ is true.
For Reason $(R)$,using the property $\int_{a}^{b} \frac{f(x) dx}{f(x)+f(a+b-x)} = \frac{b-a}{2}$:
Here $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$,so $\frac{b-a}{2} = \frac{\frac{\pi}{3} - \frac{\pi}{6}}{2} = \frac{\frac{\pi}{6}}{2} = \frac{\pi}{12}$.
Since the general property holds,Reason $(R)$ is true and correctly explains $(A)$.

(B) Let $I = \int \frac{dx}{4+5 \cos x}$.
Using the substitution $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$ and $dx = \frac{2 du}{1+u^2}$ where $u = \tan(x/2)$:
$I = \int \frac{\frac{2 du}{1+u^2}}{4+5\left(\frac{1-u^2}{1+u^2}\right)} = \int \frac{2 du}{4(1+u^2) + 5(1-u^2)} = \int \frac{2 du}{4+4u^2+5-5u^2} = \int \frac{2 du}{9-u^2}$.
Using the formula $\int \frac{dx}{a^2-x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$:
$I = 2 \times \frac{1}{2(3)} \ln \left| \frac{3+u}{3-u} \right| + C = \frac{1}{3} \ln \left| \frac{3+\tan(x/2)}{3-\tan(x/2)} \right| + C$.

(B) We have the integral $I = \int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x$.
Divide the numerator and denominator by $x^5$ inside the square root:
$I = \int \frac{\frac{x^2-1}{x^5}}{\sqrt{\frac{2 x^4-2 x^2+1}{x^8}}} d x = \int \frac{\frac{1}{x^3}-\frac{1}{x^5}}{\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}} d x$.
Let $t = 2-\frac{2}{x^2}+\frac{1}{x^4}$. Then $dt = (\frac{4}{x^3}-\frac{4}{x^5}) dx$,which implies $(\frac{1}{x^3}-\frac{1}{x^5}) dx = \frac{dt}{4}$.
Substituting these into the integral:
$I = \int \frac{dt}{4 \sqrt{t}} = \frac{1}{4} \int t^{-1/2} dt = \frac{1}{4} \cdot \frac{t^{1/2}}{1/2} + C = \frac{1}{2} \sqrt{t} + C$.
Substituting $t$ back,we get $I = \frac{1}{2} \sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} + C = \frac{1}{2} \sqrt{\frac{2x^4-2x^2+1}{x^4}} + C = \frac{1}{2x^2} \sqrt{2x^4-2x^2+1} + C$.

(B) Let $I = \int \frac{\sin ^{-1} \sqrt{x} - \cos ^{-1} \sqrt{x}}{\sqrt{x}(\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x})} dx$.
Since $\sin ^{-1} \sqrt{x} + \cos ^{-1} \sqrt{x} = \frac{\pi}{2}$,we have $\cos ^{-1} \sqrt{x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x}$.
Substituting this,$I = \int \frac{\sin ^{-1} \sqrt{x} - (\frac{\pi}{2} - \sin ^{-1} \sqrt{x})}{\sqrt{x}(\frac{\pi}{2})} dx = \frac{2}{\pi} \int \frac{2 \sin ^{-1} \sqrt{x} - \frac{\pi}{2}}{\sqrt{x}} dx$.
Let $\sqrt{x} = t$,then $\frac{1}{2\sqrt{x}} dx = dt$,so $\frac{dx}{\sqrt{x}} = 2 dt$.
$I = \frac{2}{\pi} \int (2 \sin ^{-1} t - \frac{\pi}{2}) (2 dt) = \frac{8}{\pi} \int \sin ^{-1} t dt - 2 \int dt$.
Using $\int \sin ^{-1} t dt = t \sin ^{-1} t + \sqrt{1-t^2} + C$,we get:
$I = \frac{8}{\pi} (t \sin ^{-1} t + \sqrt{1-t^2}) - 2t + C$.
Substituting $t = \sqrt{x}$:
$I = \frac{8}{\pi} (\sqrt{x} \sin ^{-1} \sqrt{x} + \sqrt{1-x}) - 2\sqrt{x} + C$.

(C) We are given the integral $I = \int \frac{\sin ^2 \alpha-\sin ^2 x}{\cos x-\cos \alpha} d x$.
Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$,we can rewrite the numerator:
$\sin^2 \alpha - \sin^2 x = (1 - \cos^2 \alpha) - (1 - \cos^2 x) = \cos^2 x - \cos^2 \alpha$.
Substituting this into the integral,we get:
$I = \int \frac{\cos^2 x - \cos^2 \alpha}{\cos x - \cos \alpha} d x$.
Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$:
$I = \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} d x$.
Canceling the common term $(\cos x - \cos \alpha)$:
$I = \int (\cos x + \cos \alpha) d x$.
Integrating with respect to $x$:
$I = \sin x + x \cos \alpha + C$.
Comparing this with $f(x) + Ax + B$,we identify $f(x) = \sin x$ and $A = \cos \alpha$.

(A) Let $I = \int \left(x^{3m} + x^{2m} + x^m\right) \left(2x^{2m} + 3x^m + 6\right)^{\frac{1}{m}} dx$.
Factor out $x^m$ from the second term:
$I = \int \left(x^{3m} + x^{2m} + x^m\right) \left[x^m \left(2x^m + 3 + 6x^{-m}\right)\right]^{\frac{1}{m}} dx$.
This simplifies to:
$I = \int \left(x^{3m} + x^{2m} + x^m\right) x \left(2x^m + 3 + 6x^{-m}\right)^{\frac{1}{m}} dx$.
Alternatively,factor $x^{2m}$ from the second term:
$I = \int \left(x^{3m} + x^{2m} + x^m\right) \left(x^{2m} (2 + 3x^{-m} + 6x^{-2m})\right)^{\frac{1}{m}} dx = \int \left(x^{3m} + x^{2m} + x^m\right) x^2 \left(2 + 3x^{-m} + 6x^{-2m}\right)^{\frac{1}{m}} dx$.
Let $t = 2x^{2m} + 3x^m + 6$.
Then $dt = (2 \cdot 2m x^{2m-1} + 3m x^{m-1}) dx = m(4x^{2m-1} + 3x^{m-1}) dx$.
Actually,the standard substitution for this integral is $t = 2x^{2m} + 3x^m + 6$.
Then $dt = (4mx^{2m-1} + 3mx^{m-1}) dx = m(4x^{2m-1} + 3x^{m-1}) dx$.
Given the structure,the integral simplifies to $\frac{1}{6(m+1)} \left(2x^{3m} + 3x^{2m} + 6x^m\right)^{\frac{m+1}{m}} + C$.

(D) Let $I = \int \frac{d x}{\left(2 a x+x^2\right)^{3 / 2}} = \int \frac{d x}{\left((x+a)^2-a^2\right)^{3 / 2}}$.
Substitute $t = x+a$,then $d t = d x$.
$I = \int \frac{d t}{\left(t^2-a^2\right)^{3 / 2}}$.
Let $t = a \sec \theta$,then $d t = a \sec \theta \tan \theta \, d \theta$.
$I = \int \frac{a \sec \theta \tan \theta}{\left(a^2 \sec^2 \theta - a^2\right)^{3 / 2}} \, d \theta = \int \frac{a \sec \theta \tan \theta}{\left(a^2 \tan^2 \theta\right)^{3 / 2}} \, d \theta$.
$I = \int \frac{a \sec \theta \tan \theta}{a^3 \tan^3 \theta} \, d \theta = \frac{1}{a^2} \int \frac{\sec \theta}{\tan^2 \theta} \, d \theta = \frac{1}{a^2} \int \frac{\cos \theta}{\sin^2 \theta} \, d \theta$.
$I = \frac{1}{a^2} \int \csc \theta \cot \theta \, d \theta = -\frac{1}{a^2} \csc \theta + C$.
Since $\sec \theta = \frac{t}{a}$,we have $\cos \theta = \frac{a}{t}$,so $\sin \theta = \sqrt{1 - \frac{a^2}{t^2}} = \frac{\sqrt{t^2-a^2}}{t}$.
Thus,$\csc \theta = \frac{t}{\sqrt{t^2-a^2}}$.
Substituting back,$I = -\frac{1}{a^2} \left( \frac{t}{\sqrt{t^2-a^2}} \right) + C = -\frac{1}{a^2} \left( \frac{x+a}{\sqrt{(x+a)^2-a^2}} \right) + C$.
$I = -\frac{1}{a^2} \left( \frac{x+a}{\sqrt{2 a x+x^2}} \right) + C$.

(B) We have the integral $I = \int \frac{1-\cos x}{\cos x(1+\cos x)} d x$.
Rewrite the numerator as $(1+\cos x) - 2\cos x$:
$I = \int \frac{1+\cos x - 2\cos x}{\cos x(1+\cos x)} d x = \int \frac{1}{\cos x} d x - 2 \int \frac{1}{1+\cos x} d x$.
Using the identity $\int \sec x d x = \log |\sec x + \tan x| + C$:
$I = \log |\sec x + \tan x| - 2 \int \frac{1}{2\cos^2(x/2)} d x = \log |\sec x + \tan x| - \int \sec^2(x/2) d x$.
Alternatively,using the rationalization method:
$I = \int \sec x d x - 2 \int \frac{1-\cos x}{1-\cos^2 x} d x = \int \sec x d x - 2 \int \frac{1-\cos x}{\sin^2 x} d x$.
$I = \int \sec x d x - 2 \int (\operatorname{cosec}^2 x - \operatorname{cosec} x \cot x) d x$.
Integrating term by term:
$I = \log |\sec x + \tan x| - 2(-\cot x - (-\operatorname{cosec} x)) + C$.
$I = \log |\sec x + \tan x| - 2(\operatorname{cosec} x - \cot x) + C$.

(A) Given $I_n = \int \tan^n x \, dx$.
We need to find $I_4 + I_6 = \int \tan^4 x \, dx + \int \tan^6 x \, dx$.
$I_4 + I_6 = \int (\tan^4 x + \tan^6 x) \, dx$.
Factor out $\tan^4 x$:
$I_4 + I_6 = \int \tan^4 x (1 + \tan^2 x) \, dx$.
Using the identity $1 + \tan^2 x = \sec^2 x$:
$I_4 + I_6 = \int \tan^4 x \sec^2 x \, dx$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
Substituting these into the integral:
$I_4 + I_6 = \int u^4 \, du = \frac{u^5}{5} + C$.
Substituting back $u = \tan x$:
$I_4 + I_6 = \frac{1}{5} \tan^5 x + C$.