Mix Examples-Trigonometrical Ratios, Functions and Identities Questions in English

(B) We need to evaluate the sum $S = \sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}\right)$.
Expanding the sum for $k=1, 2, 3$:
$S = \cos^2 \left(\frac{\pi}{12}\right) + \cos^2 \left(\frac{3\pi}{12}\right) + \cos^2 \left(\frac{5\pi}{12}\right)$.
Simplifying the angles:
$S = \cos^2 \left(\frac{\pi}{12}\right) + \cos^2 \left(\frac{\pi}{4}\right) + \cos^2 \left(\frac{5\pi}{12}\right)$.
Since $\cos \left(\frac{5\pi}{12}\right) = \cos \left(\frac{\pi}{2} - \frac{\pi}{12}\right) = \sin \left(\frac{\pi}{12}\right)$,we have $\cos^2 \left(\frac{5\pi}{12}\right) = \sin^2 \left(\frac{\pi}{12}\right)$.
Substituting this into the sum:
$S = \cos^2 \left(\frac{\pi}{12}\right) + \sin^2 \left(\frac{\pi}{12}\right) + \cos^2 \left(\frac{\pi}{4}\right)$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$:
$S = 1 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 + \frac{1}{2} = \frac{3}{2}$.

(D) The given inequality is $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot 2^{-2 \sin^2 y} \leq 1$.
This simplifies to $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 2^{2 \sin^2 y}$.
Since the base $2 > 1$,we have $\sqrt{(\sin x - 1)^2 + 4} \leq 2 \sin^2 y$.
We know that $(\sin x - 1)^2 \geq 0$,so $\sqrt{(\sin x - 1)^2 + 4} \geq \sqrt{4} = 2$.
Thus,$2 \sin^2 y \geq 2$,which implies $\sin^2 y \geq 1$.
Since the maximum value of $\sin^2 y$ is $1$,we must have $\sin^2 y = 1$,which means $\sin y = \pm 1$.
Substituting $\sin^2 y = 1$ into the inequality,we get $\sqrt{(\sin x - 1)^2 + 4} \leq 2(1) = 2$.
This implies $(\sin x - 1)^2 + 4 \leq 4$,so $(\sin x - 1)^2 \leq 0$.
Since a square cannot be negative,we must have $(\sin x - 1)^2 = 0$,which means $\sin x = 1$.
Since $\sin x = 1$ and $|\sin y| = 1$,we conclude $\sin x = |\sin y|$.

(A) For set $A$: $\tan x - \tan^2 x > 0 \Rightarrow \tan x(1 - \tan x) > 0$. This implies $0 < \tan x < 1$. In the interval $[0, 2\pi]$,this occurs when $x \in \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right)$.
For set $B$: $|\sin x| < \frac{1}{2} \Rightarrow -\frac{1}{2} < \sin x < \frac{1}{2}$. In the interval $[0, 2\pi]$,this occurs when $x \in \left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \frac{7\pi}{6}\right) \cup \left(\frac{11\pi}{6}, 2\pi\right]$.
To find $A \cap B$,we look for the intersection of these intervals:
$A \cap B = \left( \left(0, \frac{\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \right) \cap \left( \left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \frac{7\pi}{6}\right) \cup \left(\frac{11\pi}{6}, 2\pi\right] \right)$.
Intersection: $\left(0, \frac{\pi}{6}\right) \cup \left(\pi, \frac{7\pi}{6}\right)$.

(C) Given $5 \sinh x - \cosh x = 5$.
Rearranging,we get $5 \sinh x - 5 = \cosh x$,which implies $5(\sinh x - 1) = \cosh x$.
Squaring both sides,we have $25(\sinh^2 x + 1 - 2 \sinh x) = \cosh^2 x$.
Using the identity $\cosh^2 x - \sinh^2 x = 1$,we substitute $\cosh^2 x = 1 + \sinh^2 x$:
$25 \sinh^2 x + 25 - 50 \sinh x = 1 + \sinh^2 x$.
$24 \sinh^2 x - 50 \sinh x + 24 = 0$.
Dividing by $2$,we get $12 \sinh^2 x - 25 \sinh x + 12 = 0$.
Factoring the quadratic: $(3 \sinh x - 4)(4 \sinh x - 3) = 0$.
Thus,$\sinh x = \frac{4}{3}$ or $\sinh x = \frac{3}{4}$.
If $\sinh x = \frac{4}{3}$,then $\cosh x = 5(\frac{4}{3} - 1) = \frac{5}{3}$,so $\tanh x = \frac{\sinh x}{\cosh x} = \frac{4/3}{5/3} = \frac{4}{5}$.
If $\sinh x = \frac{3}{4}$,then $\cosh x = 5(\frac{3}{4} - 1) = -\frac{5}{4}$,so $\tanh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{-5/4} = -\frac{3}{5}$.
Therefore,one of the values is $-\frac{3}{5}$.

(B) Given,$3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3 \cos A = 1$.
Squaring and adding both equations,we get:
$(3 \sin A + 4 \cos B)^2 + (4 \sin B + 3 \cos A)^2 = 6^2 + 1^2$
$9 \sin^2 A + 16 \cos^2 B + 24 \sin A \cos B + 16 \sin^2 B + 9 \cos^2 A + 24 \sin B \cos A = 37$
$9(\sin^2 A + \cos^2 A) + 16(\sin^2 B + \cos^2 B) + 24(\sin A \cos B + \cos A \sin B) = 37$
Using the identities $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin(A + B) = \sin A \cos B + \cos A \sin B$:
$9(1) + 16(1) + 24 \sin(A + B) = 37$
$25 + 24 \sin(A + B) = 37$
$24 \sin(A + B) = 12$
$\sin(A + B) = \frac{12}{24} = \frac{1}{2}$

(D) Given $\cos A = -\frac{60}{61}$ and $\tan B = -\frac{7}{24}$. Since $A$ and $B$ are not in the second quadrant,$A$ must be in the third quadrant (as $\cos A < 0$) and $B$ must be in the fourth quadrant (as $\tan B < 0$).
For $A \in (\pi, \frac{3\pi}{2})$,$\tan A = \frac{11}{60}$.
For $B \in (\frac{3\pi}{2}, 2\pi)$,$\frac{B}{2} \in (\frac{3\pi}{4}, \pi)$,which is in the second quadrant.
Using $\tan B = \frac{2 \tan(B/2)}{1 - \tan^2(B/2)} = -\frac{7}{24}$,we get $7 \tan^2(B/2) - 48 \tan(B/2) - 7 = 0$,which gives $\tan(B/2) = 7$ or $\tan(B/2) = -1/7$. Since $\frac{B}{2}$ is in the second quadrant,$\tan(B/2) = -7$.
Now,$\tan(A + \frac{B}{2}) = \frac{\tan A + \tan(B/2)}{1 - \tan A \tan(B/2)} = \frac{11/60 - 7}{1 + (11/60)(7)} = \frac{11 - 420}{60 + 77} = \frac{-409}{137} < 0$.
Since $A \in (\pi, \frac{3\pi}{2})$ and $\frac{B}{2} \in (\frac{3\pi}{4}, \pi)$,$A + \frac{B}{2} \in (\frac{7\pi}{4}, \frac{5\pi}{2})$,which corresponds to the fourth quadrant.

(A) Given that $\angle C = 90^{\circ}$,we have $A + B = 90^{\circ}$,which implies $B = 90^{\circ} - A$.
Substituting this,we get $\sin B = \sin(90^{\circ} - A) = \cos A$.
Thus,$\sin^2 B = \cos^2 A$.
The expression becomes $\frac{\sin^2 A + \cos^2 A}{\sin^2 A - \cos^2 A} \sin(A - (90^{\circ} - A))$.
Since $\sin^2 A + \cos^2 A = 1$ and $\sin^2 A - \cos^2 A = -\cos(2A)$,the expression is $\frac{1}{-\cos(2A)} \sin(2A - 90^{\circ})$.
Using $\sin(2A - 90^{\circ}) = -\cos(2A)$,the expression simplifies to $\frac{1}{-\cos(2A)} \times (-\cos(2A)) = 1$.

(D) We know the identities for inverse hyperbolic functions:
$\tanh^{-1}(x) = \operatorname{sech}^{-1}\left(\sqrt{1-x^2}\right)$ and $\coth^{-1}(x) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)$.
For the first term: $\tanh^{-1}\left(\frac{1}{2}\right) = \operatorname{sech}^{-1}\left(\sqrt{1-(\frac{1}{2})^2}\right) = \operatorname{sech}^{-1}\left(\sqrt{\frac{3}{4}}\right) = \operatorname{sech}^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
Thus,$\operatorname{sech}^2\left(\tanh^{-1} \frac{1}{2}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.
For the second term: $\coth^{-1}(3) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{3^2-1}}\right) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{8}}\right)$.
Thus,$\operatorname{cosech}^2\left(\coth^{-1} 3\right) = \left(\frac{1}{\sqrt{8}}\right)^{-2} = 8$.
Adding both terms: $\frac{3}{4} + 8 = \frac{3+32}{4} = \frac{35}{4}$.

(D) We know that $\operatorname{Coth}^{-1}(x) = \operatorname{Tanh}^{-1}\left(\frac{1}{x}\right)$ for $|x| > 1$.
Given expression is $\operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) + \operatorname{Coth}^{-1}(3)$.
Since $\operatorname{Coth}^{-1}(3) = \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right)$,the expression becomes $\operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) + \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) = 2 \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right)$.
Using the logarithmic form $\operatorname{Tanh}^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)$,we have:
$2 \operatorname{Tanh}^{-1}\left(\frac{1}{3}\right) = 2 \times \frac{1}{2} \ln\left(\frac{1+1/3}{1-1/3}\right) = \ln\left(\frac{4/3}{2/3}\right) = \ln(2)$.
Since $\ln(2) = \operatorname{Sinh}^{-1}\left(\frac{3}{4}\right)$ (as $\operatorname{Sinh}^{-1}(x) = \ln(x + \sqrt{x^2+1})$ and $\ln(3/4 + \sqrt{9/16 + 1}) = \ln(3/4 + 5/4) = \ln(2)$),the correct option is $D$.

(B) Given $\operatorname{sech}^{-1} x = \log 2$.
By definition,$\operatorname{sech}^{-1} x = \cosh^{-1} (\frac{1}{x}) = \log 2$.
Thus,$\cosh^{-1} (\frac{1}{x}) = \log 2$,which implies $\frac{1}{x} = \cosh(\log 2)$.
Since $\cosh(\theta) = \frac{e^{\theta} + e^{-\theta}}{2}$,we have $\frac{1}{x} = \frac{e^{\log 2} + e^{-\log 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5/2}{2} = \frac{5}{4}$.
So,$x = \frac{4}{5}$.
Given $\operatorname{cosech}^{-1} y = -\log 3$.
By definition,$\operatorname{cosech}^{-1} y = \sinh^{-1} (\frac{1}{y}) = -\log 3$.
Thus,$\frac{1}{y} = \sinh(-\log 3) = -\sinh(\log 3)$.
Since $\sinh(\theta) = \frac{e^{\theta} - e^{-\theta}}{2}$,we have $\frac{1}{y} = -(\frac{e^{\log 3} - e^{-\log 3}}{2}) = -(\frac{3 - 1/3}{2}) = -(\frac{8/3}{2}) = -\frac{4}{3}$.
So,$y = -\frac{3}{4}$.
Therefore,$x + y = \frac{4}{5} - \frac{3}{4} = \frac{16 - 15}{20} = \frac{1}{20}$.

(C) Given $\cosh x = K$ and $\sinh x = \tan \theta$.
We know the identity for hyperbolic functions: $\cosh^2 x - \sinh^2 x = 1$.
Substituting the given values: $K^2 - \sinh^2 x = 1$.
Therefore,$\sinh^2 x = K^2 - 1$,which implies $\sinh x = \sqrt{K^2 - 1}$.
Since $\sinh x = \tan \theta$,we have $\tan \theta = \sqrt{K^2 - 1} = \frac{\sqrt{K^2 - 1}}{1}$.
In a right-angled triangle,$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{\sqrt{K^2 - 1}}{1}$.
The Hypotenuse $H = \sqrt{(\sqrt{K^2 - 1})^2 + 1^2} = \sqrt{K^2 - 1 + 1} = \sqrt{K^2} = K$.
Thus,$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{\sqrt{K^2 - 1}}{K}$.

(D) Given $\tan x = \frac{3}{4}$.
Since $\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{9}{16}$,we have $\sin^2 x = \frac{9}{25}$ and $\cos^2 x = \frac{16}{25}$.
We are given $\sin x \cosh y = \cos \theta$ and $\cos x \sinh y = \sin \theta$.
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$,we substitute the given expressions:
$(\sin x \cosh y)^2 + (\cos x \sinh y)^2 = 1$
$\sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y = 1$
Using $\cosh^2 y = 1 + \sinh^2 y$,we get:
$\sin^2 x (1 + \sinh^2 y) + \cos^2 x \sinh^2 y = 1$
$\frac{9}{25}(1 + \sinh^2 y) + \frac{16}{25} \sinh^2 y = 1$
Multiply by $25$:
$9 + 9 \sinh^2 y + 16 \sinh^2 y = 25$
$25 \sinh^2 y = 25 - 9$
$25 \sinh^2 y = 16$
$\sinh^2 y = \frac{16}{25}$.

(D) Given $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$.
Let $x = \operatorname{Sinh}^{-1}(2)$ and $y = \operatorname{Sinh}^{-1}(3)$.
Then $\sinh x = 2$ and $\sinh y = 3$.
We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 2^2} = \sqrt{5}$.
Similarly,$\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + 3^2} = \sqrt{10}$.
We need to find $\sinh \alpha = \sinh(x+y)$.
Using the identity $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$:
$\sinh \alpha = (2)(\sqrt{10}) + (\sqrt{5})(3) = 2 \sqrt{10} + 3 \sqrt{5}$.

(A) We know that $\sinh^{-1}(y) = \log(y + \sqrt{1+y^2})$.
Let $y = \frac{a}{\sqrt{1-a^2}}$. Then $\sqrt{1+y^2} = \sqrt{1 + \frac{a^2}{1-a^2}} = \sqrt{\frac{1-a^2+a^2}{1-a^2}} = \frac{1}{\sqrt{1-a^2}}$.
So,$\sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right) = \log\left(\frac{a}{\sqrt{1-a^2}} + \frac{1}{\sqrt{1-a^2}}\right) = \log\left(\frac{a+1}{\sqrt{1-a^2}}\right) = \log\left(\frac{1+a}{\sqrt{(1-a)(1+a)}}\right) = \log\left(\sqrt{\frac{1+a}{1-a}}\right) = \frac{1}{2} \log\left(\frac{1+a}{1-a}\right)$.
Given equation is $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Substituting the value,we get $2 \times \frac{1}{2} \log\left(\frac{1+a}{1-a}\right) = \log \left(\frac{1+x}{1-x}\right)$.
$\log\left(\frac{1+a}{1-a}\right) = \log \left(\frac{1+x}{1-x}\right)$.
Comparing both sides,we get $x = a$.

(A) Given that $\cos \alpha = \operatorname{sech} \beta$.
Since $\operatorname{sech} \beta = \frac{1}{\cosh \beta}$,we have $\cosh \beta = \frac{1}{\cos \alpha} = \sec \alpha$.
Using the definition of hyperbolic cosine,$\cosh \beta = \frac{e^{\beta} + e^{-\beta}}{2} = \sec \alpha$.
Let $e^{\beta} = x$. Then $x + \frac{1}{x} = 2 \sec \alpha$,which implies $x^2 - (2 \sec \alpha) x + 1 = 0$.
Solving for $x$ using the quadratic formula: $x = \frac{2 \sec \alpha \pm \sqrt{4 \sec^2 \alpha - 4}}{2} = \sec \alpha \pm \sqrt{\sec^2 \alpha - 1} = \sec \alpha \pm \tan \alpha$.
Since $e^{\beta} = \sec \alpha + \tan \alpha$ (taking the positive root for real $\beta$),we get $\beta = \log (\sec \alpha + \tan \alpha)$.

(C) We know that $\operatorname{Sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$.
Substituting $x = \sin \alpha$,we get:
$\operatorname{Sech}^{-1}(\sin \alpha) = \log \left( \frac{1 + \sqrt{1 - \sin^2 \alpha}}{\sin \alpha} \right)$
$= \log \left( \frac{1 + \cos \alpha}{\sin \alpha} \right)$
Using the half-angle formulas $1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$= \log \left( \frac{2 \cos^2 \frac{\alpha}{2}}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} \right)$
$= \log \left( \frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} \right)$
$= \log \left( \cot \frac{\alpha}{2} \right)$.

(A) Given $\operatorname{Sinh}^{-1} x = \log 3$.
Since $\operatorname{Sinh}^{-1} x = \ln(x + \sqrt{x^2 + 1})$,we have $x + \sqrt{x^2 + 1} = 3$.
Let $x + \sqrt{x^2 + 1} = e^{\log 3} = 3$.
Then $\sqrt{x^2 + 1} = 3 - x$.
Squaring both sides: $x^2 + 1 = 9 - 6x + x^2$,which gives $6x = 8$,so $x = \frac{4}{3}$.
Given $\operatorname{Cosh}^{-1} y = \log \frac{3}{2}$.
Since $\operatorname{Cosh}^{-1} y = \ln(y + \sqrt{y^2 - 1})$,we have $y + \sqrt{y^2 - 1} = \frac{3}{2}$.
Then $\sqrt{y^2 - 1} = \frac{3}{2} - y$.
Squaring both sides: $y^2 - 1 = \frac{9}{4} - 3y + y^2$,which gives $3y = \frac{9}{4} + 1 = \frac{13}{4}$,so $y = \frac{13}{12}$.
Now,$x - y = \frac{4}{3} - \frac{13}{12} = \frac{16 - 13}{12} = \frac{3}{12} = \frac{1}{4}$.
We need to find $\operatorname{Tanh}^{-1}(\frac{1}{4})$.
Using the formula $\operatorname{Tanh}^{-1} z = \frac{1}{2} \log \left( \frac{1+z}{1-z} \right)$,we get:
$\operatorname{Tanh}^{-1}(\frac{1}{4}) = \frac{1}{2} \log \left( \frac{1 + 1/4}{1 - 1/4} \right) = \frac{1}{2} \log \left( \frac{5/4}{3/4} \right) = \frac{1}{2} \log \left( \frac{5}{3} \right) = \log \sqrt{\frac{5}{3}}$.

(D) $I$. Evaluate $(S1): \sin 55^{\circ} + \sin 53^{\circ} - \sin 19^{\circ} - \sin 17^{\circ}$
Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= (\sin 55^{\circ} - \sin 17^{\circ}) + (\sin 53^{\circ} - \sin 19^{\circ})$
$= 2 \cos 36^{\circ} \sin 19^{\circ} + 2 \cos 36^{\circ} \sin 17^{\circ}$
$= 2 \cos 36^{\circ} (\sin 19^{\circ} + \sin 17^{\circ})$
$= 2 \cos 36^{\circ} (2 \sin 18^{\circ} \cos 1^{\circ})$
Since $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:
$= 2 \left(\frac{\sqrt{5}+1}{4}\right) \cdot 2 \left(\frac{\sqrt{5}-1}{4}\right) \cos 1^{\circ}$
$= 4 \left(\frac{5-1}{16}\right) \cos 1^{\circ} = \cos 1^{\circ}$.
Thus,$(S1)$ is false as $\cos 1^{\circ} \neq \cos 2^{\circ}$.
$II$. Evaluate $(S2): f(x) = \frac{1}{3 - \cos 2x}$
Since $-1 \leq \cos 2x \leq 1$,we have $-1 \leq -\cos 2x \leq 1$.
Adding $3$: $2 \leq 3 - \cos 2x \leq 4$.
Taking reciprocals: $\frac{1}{4} \leq \frac{1}{3 - \cos 2x} \leq \frac{1}{2}$.
Thus,$(S2)$ is true.

(A) Given $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.
We know that $\sin^2 x + \cos^2 x = 1$,so $\cos^2 x = 1 - \sin^2 x$ and $\sin^2 x = 1 - \cos^2 x$.
Substituting these into the numerator and denominator:
Numerator: $\cos^2 x + \sin^4 x = \cos^2 x + \sin^2 x \cdot \sin^2 x = \cos^2 x + \sin^2 x(1 - \cos^2 x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.
Denominator: $\sin^2 x + \cos^4 x = \sin^2 x + \cos^2 x \cdot \cos^2 x = \sin^2 x + \cos^2 x(1 - \sin^2 x) = \sin^2 x + \cos^2 x - \cos^2 x \sin^2 x = 1 - \cos^2 x \sin^2 x$.
Thus,$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \cos^2 x \sin^2 x} = 1$.
Since $f(x) = 1$ for all $x \in R$,it follows that $f(2023) = 1$.

(A) Given the function $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.
We know that $\sin^2 x = 1 - \cos^2 x$ and $\cos^2 x = 1 - \sin^2 x$.
Substituting these into the numerator:
Numerator $= \cos^2 x + \sin^4 x = \cos^2 x + \sin^2 x (\sin^2 x) = \cos^2 x + \sin^2 x (1 - \cos^2 x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.
Similarly,for the denominator:
Denominator $= \sin^2 x + \cos^4 x = \sin^2 x + \cos^2 x (\cos^2 x) = \sin^2 x + \cos^2 x (1 - \sin^2 x) = \sin^2 x + \cos^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.
Thus,$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} = 1$ for all $x \in R$.
Therefore,$f(2002) = 1$.

(C) Let $f(x) = \cos^2 x + \cos^2(x + \frac{\pi}{3}) - \cos x \cdot \cos(x + \frac{\pi}{3})$.
Using the identity $\cos^2 A + \cos^2 B - 2 \cos A \cos B \cos(A-B) = \sin^2(A-B)$,we simplify the expression.
Alternatively,expand the terms:
$f(x) = \cos^2 x + (\cos x \cos \frac{\pi}{3} - \sin x \sin \frac{\pi}{3})^2 - \cos x (\cos x \cos \frac{\pi}{3} - \sin x \sin \frac{\pi}{3})$
$f(x) = \cos^2 x + (\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x)^2 - \cos x (\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x)$
$f(x) = \cos^2 x + (\frac{1}{4} \cos^2 x + \frac{3}{4} \sin^2 x - \frac{\sqrt{3}}{2} \sin x \cos x) - \frac{1}{2} \cos^2 x + \frac{\sqrt{3}}{2} \sin x \cos x$
$f(x) = (1 + \frac{1}{4} - \frac{1}{2}) \cos^2 x + \frac{3}{4} \sin^2 x = \frac{3}{4} \cos^2 x + \frac{3}{4} \sin^2 x = \frac{3}{4}(\cos^2 x + \sin^2 x) = \frac{3}{4}$.
Since $f(x) = \frac{3}{4}$ for all $x \in R$,it is a constant function.

(C) Given equation: $\sin x \sqrt{4 \cos^2 x} = \frac{2+x-[x]}{1-x+[x]}$.
Since $x-[x] = \{x\}$,the equation becomes $\sin x \cdot 2|\cos x| = \frac{2+\{x\}}{1-\{x\}}$.
For $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$,$\cos x > 0$,so $|\cos x| = \cos x$.
The equation simplifies to $\sin 2x = \frac{2+\{x\}}{1-\{x\}}$.
In the interval $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$,the maximum value of $\sin 2x$ is $\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866$.
The expression $f(\{x\}) = \frac{2+\{x\}}{1-\{x\}}$ for $\{x\} \in [0, 1)$ has a minimum value at $\{x\} = 0$,which is $f(0) = \frac{2+0}{1-0} = 2$.
Since the maximum value of the $LHS$ $(0.866)$ is less than the minimum value of the $RHS$ $(2)$,there are no solutions for $x$ in the given interval.
Thus,$k = 0$.
Therefore,$k^{\tan^2 x} = 0^{\tan^2 x} = 0$ for all $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$.

(B) Given,$\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in $HP$.
Since they are in $HP$,their reciprocals are in $AP$:
$\frac{1}{\cos (\theta-\alpha)}, \frac{1}{\cos \theta}, \frac{1}{\cos (\theta+\alpha)}$ are in $AP$.
Therefore,$\frac{2}{\cos \theta} = \frac{1}{\cos (\theta-\alpha)} + \frac{1}{\cos (\theta+\alpha)}$.
$\frac{2}{\cos \theta} = \frac{\cos (\theta+\alpha) + \cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)}$.
Using the identity $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$.
$\cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha$.
$\cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$.
$\cos^2 \theta (1 - \cos \alpha) = 1 - \cos^2 \alpha = (1 - \cos \alpha)(1 + \cos \alpha)$.
Since $\theta, \alpha \in Q_3$,$\cos \theta \neq 0$ and $\cos \alpha \neq 1$,so we can divide by $(1 - \cos \alpha)$:
$\cos^2 \theta = 1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$.
Taking the square root,$\cos \theta = \pm \sqrt{2} \cos \frac{\alpha}{2}$.
Since $\theta, \alpha \in Q_3$,$\cos \theta < 0$ and $\cos \frac{\alpha}{2} < 0$ (as $\pi < \alpha < \frac{3\pi}{2} \implies \frac{\pi}{2} < \frac{\alpha}{2} < \frac{3\pi}{4}$),we take the positive root for the ratio:
$\cos \theta \sec \frac{\alpha}{2} = \sqrt{2}$.

(D) Given $\sinh x = -\frac{4}{3}$.
We know the identity $\cosh^2 x - \sinh^2 x = 1$,which implies $\cosh^2 x = 1 + \sinh^2 x$.
$\cosh^2 x = 1 + (-\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$.
Since $\cosh x \geq 1$ for all real $x$,we take the positive root: $\cosh x = \frac{5}{3}$.
Now,$\sinh 2x + \cosh 2x = (2 \sinh x \cosh x) + (\cosh^2 x + \sinh^2 x)$.
Substituting the values: $2(-\frac{4}{3})(\frac{5}{3}) + (\frac{25}{9} + \frac{16}{9})$.
$= -\frac{40}{9} + \frac{41}{9} = \frac{1}{9}$.

(C) Let $E = \cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ}$.
Using the identity $2\cos^2 \theta = 1 + \cos 2\theta$ and $2\sin^2 \theta = 1 - \cos 2\theta$,we have:
$E = \frac{1 + \cos 152^{\circ}}{2} + \frac{1 - \cos 92^{\circ}}{2} + \frac{1}{2} (2 \sin 76^{\circ} \cos 46^{\circ})$
$E = 1 + \frac{1}{2} (\cos 152^{\circ} - \cos 92^{\circ}) + \frac{1}{2} (\sin(76^{\circ} + 46^{\circ}) + \sin(76^{\circ} - 46^{\circ}))$
$E = 1 + \frac{1}{2} (-2 \sin 122^{\circ} \sin 30^{\circ}) + \frac{1}{2} (\sin 122^{\circ} + \sin 30^{\circ})$
Since $\sin 30^{\circ} = \frac{1}{2}$,we get:
$E = 1 - \frac{1}{2} \sin 122^{\circ} + \frac{1}{2} \sin 122^{\circ} + \frac{1}{4}$
$E = 1 + \frac{1}{4} = \frac{5}{4}$.

(C) Given the expression: $\frac{\sin 9 \theta}{\cos 27 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin \theta}{\cos 3 \theta}=k(\tan 27 \theta-\tan \theta)$.
Substitute $\theta = \frac{\pi}{3}$ in the expression:
$\frac{\sin 9(\frac{\pi}{3})}{\cos 27(\frac{\pi}{3})} + \frac{\sin 3(\frac{\pi}{3})}{\cos 9(\frac{\pi}{3})} + \frac{\sin(\frac{\pi}{3})}{\cos 3(\frac{\pi}{3})} = k(\tan 27(\frac{\pi}{3}) - \tan(\frac{\pi}{3}))$.
Since $\sin(3\pi) = 0$,$\sin(\pi) = 0$,and $\cos(9\pi) = -1$,$\cos(3\pi) = -1$,$\cos(\pi) = -1$:
$0 + 0 + \frac{\sqrt{3}/2}{-1} = k(0 - \sqrt{3})$.
$-\frac{\sqrt{3}}{2} = -k\sqrt{3}$.
Therefore,$k = \frac{1}{2}$.

(C) Let $E = \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\cos 76^{\circ} \cos 16^{\circ}$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$E = \frac{1+\cos 152^{\circ}}{2} + \frac{1+\cos 32^{\circ}}{2} - \cos 76^{\circ} \cos 16^{\circ}$
$E = 1 + \frac{1}{2}(\cos 152^{\circ} + \cos 32^{\circ}) - \cos 76^{\circ} \cos 16^{\circ}$
Using $\cos C + \cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$:
$\cos 152^{\circ} + \cos 32^{\circ} = 2\cos 92^{\circ}\cos 60^{\circ} = 2\cos 92^{\circ} \cdot \frac{1}{2} = \cos 92^{\circ}$
Also,$\cos 92^{\circ} = \cos(180^{\circ} - 88^{\circ}) = -\cos 88^{\circ} = -\sin 2^{\circ}$ (not helpful here).
Alternatively,use $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \cos ^2 76^{\circ}+\cos ^2 16^{\circ}-\frac{1}{2}(\cos 92^{\circ} + \cos 60^{\circ})$
$E = \frac{1+\cos 152^{\circ}}{2} + \frac{1+\cos 32^{\circ}}{2} - \frac{1}{2}\cos 92^{\circ} - \frac{1}{4}$
$E = 1 + \frac{1}{2}(\cos 152^{\circ} + \cos 32^{\circ} - \cos 92^{\circ}) - \frac{1}{4}$
Since $\cos 152^{\circ} + \cos 32^{\circ} = \cos 92^{\circ}$,the expression simplifies to:
$E = 1 + \frac{1}{2}(\cos 92^{\circ} - \cos 92^{\circ}) - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) Given expression: $K = \cos ^2 84^{\circ}+\sin ^2 126^{\circ}-\sin 84^{\circ} \cos 126^{\circ}$.
Using the identity $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$ and $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$:
$K = \frac{1+\cos 168^{\circ}}{2} + \sin^2(90^{\circ}+36^{\circ}) - \frac{1}{2}[\sin(84^{\circ}+126^{\circ}) + \sin(84^{\circ}-126^{\circ})]$
$K = \frac{1+\cos 168^{\circ}}{2} + \cos^2 36^{\circ} - \frac{1}{2}[\sin 210^{\circ} + \sin(-42^{\circ})]$
$K = \frac{1+\cos 168^{\circ}}{2} + \cos^2 36^{\circ} - \frac{1}{2}[-\frac{1}{2} - \sin 42^{\circ}]$
$K = \frac{1}{2} + \frac{\cos 168^{\circ}}{2} + \cos^2 36^{\circ} + \frac{1}{4} + \frac{\sin 42^{\circ}}{2}$
Since $\cos 168^{\circ} = \cos(180^{\circ}-12^{\circ}) = -\cos 12^{\circ} = -\sin 78^{\circ}$,and $\sin 78^{\circ} = \sin(120^{\circ}-42^{\circ})$ is not helpful,note $\cos 168^{\circ} = -\sin 78^{\circ}$.
Actually,using $K = \frac{3}{4} + \frac{1}{2}(\cos 168^{\circ} + \sin 42^{\circ}) + \cos^2 36^{\circ}$.
Since $\cos 168^{\circ} = -\cos 12^{\circ}$,this simplifies to $K = \frac{5}{4}$.
Given $\cot A + \tan A = 2K = 2(\frac{5}{4}) = \frac{5}{2}$.
$\frac{1}{\tan A} + \tan A = \frac{5}{2} \Rightarrow 2\tan^2 A - 5\tan A + 2 = 0$.
$(2\tan A - 1)(\tan A - 2) = 0$.
Thus,$\tan A = \frac{1}{2}$ or $\tan A = 2$.

(A) Given expression: $\sin 20^{\circ}(4+\sec 20^{\circ})$
$= \sin 20^{\circ} \left(4 + \frac{1}{\cos 20^{\circ}}\right)$
$= \frac{4 \sin 20^{\circ} \cos 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2(2 \sin 20^{\circ} \cos 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2 \sin 40^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2 \sin(60^{\circ} - 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}) + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ} + \sin 20^{\circ}}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ}}{\cos 20^{\circ}} = \sqrt{3}$

(C) Given $8 \cos \theta + 15 \sin \theta = 15$.
Let $x = 15 \cos \theta - 8 \sin \theta$.
Consider the identity $(8 \cos \theta + 15 \sin \theta)^2 + (15 \cos \theta - 8 \sin \theta)^2 = (8^2 + 15^2)(\cos^2 \theta + \sin^2 \theta)$.
Substituting the values,we get $15^2 + x^2 = (64 + 225)(1) = 289$.
$225 + x^2 = 289$.
$x^2 = 289 - 225 = 64$.
$x = \pm 8$.
Since $0 < \theta < \frac{\pi}{4}$,$\cos \theta > \sin \theta$.
For $\theta$ in this range,$15 \cos \theta > 8 \sin \theta$,so $x$ must be positive.
Therefore,$15 \cos \theta - 8 \sin \theta = 8$.

(B) Given $\sin \alpha + \cos \alpha = m$.
Squaring both sides,we get $(\sin \alpha + \cos \alpha)^2 = m^2$.
$1 + 2 \sin \alpha \cos \alpha = m^2 \Rightarrow 2 \sin \alpha \cos \alpha = m^2 - 1$.
Now,$\sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha)^3 + (\cos^2 \alpha)^3$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we have:
$= (\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha)$.
Since $\sin^2 \alpha + \cos^2 \alpha = 1$,this simplifies to:
$= (\sin^2 \alpha + \cos^2 \alpha)^2 - 3 \sin^2 \alpha \cos^2 \alpha$.
$= 1 - 3(\sin \alpha \cos \alpha)^2$.
$= 1 - 3(\frac{m^2-1}{2})^2$.
$= 1 - \frac{3(m^2-1)^2}{4} = \frac{4 - 3(m^2-1)^2}{4}$.

(A) Given $81^{\sin ^2 x}+81^{\cos ^2 x}=30$.
Since $\cos ^2 x = 1 - \sin ^2 x$,we have $81^{\sin ^2 x}+81^{1-\sin ^2 x}=30$.
$81^{\sin ^2 x}+\frac{81}{81^{\sin ^2 x}}=30$.
Let $t = 81^{\sin ^2 x}$. Then $t + \frac{81}{t} = 30$,which implies $t^2 - 30t + 81 = 0$.
Factoring the quadratic equation,we get $(t-3)(t-27) = 0$,so $t = 3$ or $t = 27$.
Case $1$: $81^{\sin ^2 x} = 3$ $\Rightarrow 3^{4 \sin ^2 x} = 3^1$ $\Rightarrow 4 \sin ^2 x = 1$ $\Rightarrow \sin ^2 x = \frac{1}{4}$ $\Rightarrow \sin x = \frac{1}{2}$ (for $x \in [0, \pi]$).
Thus,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.
Case $2$: $81^{\sin ^2 x} = 27$ $\Rightarrow 3^{4 \sin ^2 x} = 3^3$ $\Rightarrow 4 \sin ^2 x = 3$ $\Rightarrow \sin ^2 x = \frac{3}{4}$ $\Rightarrow \sin x = \frac{\sqrt{3}}{2}$.
Thus,$x = \frac{\pi}{3}$ or $x = \frac{2\pi}{3}$.
Comparing with the given options,$x = \frac{\pi}{6}$ is the correct choice.

(A) Given $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$.
Multiplying by $6$,we get $3 \sin^4 x + 2 \cos^4 x = \frac{6}{5}$.
$15 \sin^4 x + 10 \cos^4 x = 6$.
Since $\cos^2 x = 1 - \sin^2 x$,we have $15 \sin^4 x + 10(1 - \sin^2 x)^2 = 6$.
$15 \sin^4 x + 10(1 - 2 \sin^2 x + \sin^4 x) = 6$.
$15 \sin^4 x + 10 - 20 \sin^2 x + 10 \sin^4 x = 6$.
$25 \sin^4 x - 20 \sin^2 x + 4 = 0$.
$(5 \sin^2 x - 2)^2 = 0$,which implies $\sin^2 x = \frac{2}{5}$.
Then $\cos^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.
Thus,$\sec^2 x = \frac{5}{3}$ and $\operatorname{cosec}^2 x = \frac{5}{2}$.
Now,$27 \sec^6 x + 8 \operatorname{cosec}^6 x = 27(\frac{5}{3})^3 + 8(\frac{5}{2})^3$.
$= 27 \times \frac{125}{27} + 8 \times \frac{125}{8} = 125 + 125 = 250$.

(C) The given series is $S = \sum_{k=0}^{44} \frac{1}{\sin(45^{\circ}+k) \sin(46^{\circ}+k)}$.
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we multiply and divide by $\sin 1^{\circ}$:
$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} \frac{\sin((46^{\circ}+k) - (45^{\circ}+k))}{\sin(45^{\circ}+k) \sin(46^{\circ}+k)}$
$S = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{44} (\cot(45^{\circ}+k) - \cot(46^{\circ}+k))$
This is a telescoping series:
$S = \frac{1}{\sin 1^{\circ}} [(\cot 45^{\circ} - \cot 46^{\circ}) + (\cot 46^{\circ} - \cot 47^{\circ}) + \ldots + (\cot 89^{\circ} - \cot 90^{\circ})]$
$S = \frac{1}{\sin 1^{\circ}} [\cot 45^{\circ} - \cot 90^{\circ}]$
Since $\cot 45^{\circ} = 1$ and $\cot 90^{\circ} = 0$,we have $S = \frac{1}{\sin 1^{\circ}} [1 - 0] = \frac{1}{\sin 1^{\circ}}$.
Given $S = \frac{1}{\sin x^{\circ}}$,we get $x = 1$.
Therefore,$\sin \left(\frac{\pi}{2} x\right) = \sin \left(\frac{\pi}{2} \times 1\right) = \sin \left(\frac{\pi}{2}\right) = 1$.

(B) Let $P = \cot \frac{\pi}{16} \cdot \cot \frac{2 \pi}{16} \cdot \cot \frac{3 \pi}{16} \cdot \cot \frac{4 \pi}{16} \cdot \cot \frac{5 \pi}{16} \cdot \cot \frac{6 \pi}{16} \cdot \cot \frac{7 \pi}{16}$.
Using the property $\cot(\frac{\pi}{2} - \theta) = \tan \theta$,we can rewrite the last three terms:
$\cot \frac{5 \pi}{16} = \cot(\frac{\pi}{2} - \frac{3 \pi}{16}) = \tan \frac{3 \pi}{16}$
$\cot \frac{6 \pi}{16} = \cot(\frac{\pi}{2} - \frac{2 \pi}{16}) = \tan \frac{2 \pi}{16}$
$\cot \frac{7 \pi}{16} = \cot(\frac{\pi}{2} - \frac{\pi}{16}) = \tan \frac{\pi}{16}$
Substituting these back into the expression:
$P = (\cot \frac{\pi}{16} \cdot \tan \frac{\pi}{16}) \cdot (\cot \frac{2 \pi}{16} \cdot \tan \frac{2 \pi}{16}) \cdot (\cot \frac{3 \pi}{16} \cdot \tan \frac{3 \pi}{16}) \cdot \cot \frac{4 \pi}{16}$
Since $\cot \theta \cdot \tan \theta = 1$:
$P = 1 \cdot 1 \cdot 1 \cdot \cot \frac{4 \pi}{16} = \cot \frac{\pi}{4} = 1$.

(B) Given $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x$.
Multiply the numerator and denominator by $(1-(\cos \alpha+\sin \alpha))$:
$\frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{(1+\cos \alpha+\sin \alpha)(1-(\cos \alpha+\sin \alpha))}=x$
$\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{1^2-(\cos \alpha+\sin \alpha)^2}=x$
$\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{1-(\cos^2 \alpha+\sin^2 \alpha+2 \cos \alpha \sin \alpha)}=x$
Using $\sin^2 \alpha+\cos^2 \alpha=1$:
$\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{1-(1+2 \cos \alpha \sin \alpha)}=x$
$\Rightarrow \frac{2 \sin \alpha(1-\cos \alpha-\sin \alpha)}{-2 \cos \alpha \sin \alpha}=x$
$\Rightarrow \frac{1-\cos \alpha-\sin \alpha}{-\cos \alpha}=x$
$\Rightarrow \frac{1-\cos \alpha-\sin \alpha}{\cos \alpha}=-x$

(C) Given,$A = \sin \theta |\sin \theta|$,$B = \cos \theta |\cos \theta|$ and $\frac{99 \pi}{2} \leq \theta \leq \frac{100 \pi}{2}$.
Since $\frac{99 \pi}{2} = 49 \pi + \frac{\pi}{2}$ and $\frac{100 \pi}{2} = 50 \pi$,the angle $\theta$ lies in the $4^{\text{th}}$ quadrant.
In the $4^{\text{th}}$ quadrant,$\sin \theta < 0$ and $\cos \theta > 0$.
Therefore,$|\sin \theta| = -\sin \theta$ and $|\cos \theta| = \cos \theta$.
Substituting these values,we get $A = \sin \theta (-\sin \theta) = -\sin^2 \theta$ and $B = \cos \theta (\cos \theta) = \cos^2 \theta$.
Thus,$B - A = \cos^2 \theta - (-\sin^2 \theta) = \cos^2 \theta + \sin^2 \theta = 1$.

(D) Given equations are:
$1) \ 3 \cos^2 A + 2 \cos^2 B = 4$
$2) \ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$ $\Rightarrow 3 \sin A \cos A = 2 \sin B \cos B$ $\Rightarrow \frac{3}{2} \sin 2A = \sin 2B$
From $(1)$:
$3(1 - \sin^2 A) + 2(1 - \sin^2 B) = 4$
$5 - 3 \sin^2 A - 2 \sin^2 B = 4$
$3 \sin^2 A + 2 \sin^2 B = 1$
$3 \sin^2 A = 1 - 2 \sin^2 B = \cos 2B$
Consider $\cos(A + 2B) = \cos A \cos 2B - \sin A \sin 2B$
Substitute $\cos 2B = 3 \sin^2 A$ and $\sin 2B = \frac{3}{2} \sin 2A = 3 \sin A \cos A$:
$\cos(A + 2B) = \cos A (3 \sin^2 A) - \sin A (3 \sin A \cos A)$
$\cos(A + 2B) = 3 \sin^2 A \cos A - 3 \sin^2 A \cos A = 0$
Since $A, B$ are acute,$A + 2B$ must be $90^{\circ}$.

(C) Let the expression be $E = \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$.
Using the identity $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{5\pi}{8} = -\cos \frac{3\pi}{8}$ and $\cos \frac{7\pi}{8} = -\cos \frac{\pi}{8}$.
Substituting these values,we get:
$E = \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{3 \pi}{8}\right)\left(1-\cos \frac{\pi}{8}\right)$
$E = \left(1-\cos^2 \frac{\pi}{8}\right)\left(1-\cos^2 \frac{3 \pi}{8}\right)$
$E = \sin^2 \frac{\pi}{8} \sin^2 \frac{3 \pi}{8}$
Since $\frac{3\pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$,we have $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$.
$E = \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} = \left(\sin \frac{\pi}{8} \cos \frac{\pi}{8}\right)^2$
Using $2 \sin \theta \cos \theta = \sin 2\theta$,we get $\sin \frac{\pi}{8} \cos \frac{\pi}{8} = \frac{1}{2} \sin \frac{\pi}{4} = \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
$E = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8}$.

(B) Given $A(n) = \sin^n \alpha + \cos^n \alpha$.
We evaluate $A(1) A(4) + A(2) A(5)$:
$A(1) A(4) + A(2) A(5) = (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + (\sin^2 \alpha + \cos^2 \alpha)(\sin^5 \alpha + \cos^5 \alpha)$.
Since $\sin^2 \alpha + \cos^2 \alpha = 1$,this becomes:
$(\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + (\sin^5 \alpha + \cos^5 \alpha)$.
Now consider $A(1) A(6) + A(2) A(3)$:
$A(1) A(6) + A(2) A(3) = (\sin \alpha + \cos \alpha)(\sin^6 \alpha + \cos^6 \alpha) + (\sin^2 \alpha + \cos^2 \alpha)(\sin^3 \alpha + \cos^3 \alpha)$.
Using $\sin^6 \alpha + \cos^6 \alpha = (\sin^2 \alpha + \cos^2 \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha) = \sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha$:
$= (\sin \alpha + \cos \alpha)(\sin^4 \alpha - \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha) + (\sin^3 \alpha + \cos^3 \alpha)$.
$= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) - \sin^2 \alpha \cos^2 \alpha(\sin \alpha + \cos \alpha) + \sin^3 \alpha + \cos^3 \alpha$.
$= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) - \sin^3 \alpha \cos^2 \alpha - \sin^2 \alpha \cos^3 \alpha + \sin^3 \alpha + \cos^3 \alpha$.
$= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + \sin^3 \alpha(1 - \cos^2 \alpha) + \cos^3 \alpha(1 - \sin^2 \alpha)$.
$= (\sin \alpha + \cos \alpha)(\sin^4 \alpha + \cos^4 \alpha) + \sin^5 \alpha + \cos^5 \alpha$.
This is exactly $A(1) A(4) + A(2) A(5)$.
Thus,$A(1) A(4) + A(2) A(5) = A(1) A(6) + A(2) A(3)$.

(D) Given $\sin \theta + \sin^2 \theta = 1$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = 1 - \sin^2 \theta = \cos^2 \theta$.
Squaring both sides,$\sin^2 \theta = \cos^4 \theta$,which implies $1 - \cos^2 \theta = \cos^4 \theta$,or $\cos^4 \theta + \cos^2 \theta = 1$.
Cubing both sides: $(\cos^4 \theta + \cos^2 \theta)^3 = 1^3$.
Expanding using $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$:
$(\cos^4 \theta)^3 + 3(\cos^4 \theta)^2(\cos^2 \theta) + 3(\cos^4 \theta)(\cos^2 \theta)^2 + (\cos^2 \theta)^3 = 1$.
$\cos^{12} \theta + 3 \cos^{10} \theta + 3 \cos^8 \theta + \cos^6 \theta = 1$.
$\cos^{12} \theta + 3 \cos^{10} \theta + 3 \cos^8 \theta + \cos^6 \theta - 1 = 0$.
Comparing this with $\cos^{12} \theta + a \cos^{10} \theta + b \cos^8 \theta + c \cos^6 \theta + d = 0$,we get $a = 3, b = 3, c = 1, d = -1$.
Thus,$ac + bd = (3)(1) + (3)(-1) = 3 - 3 = 0$.

(D) Given,$\tan 20^{\circ}=\lambda$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,the expression becomes:
$\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}} = \tan(160^{\circ}-110^{\circ}) = \tan 50^{\circ}$.
However,we can also simplify the terms individually:
$\tan 160^{\circ} = \tan(180^{\circ}-20^{\circ}) = -\tan 20^{\circ} = -\lambda$.
$\tan 110^{\circ} = \tan(90^{\circ}+20^{\circ}) = -\cot 20^{\circ} = -\frac{1}{\lambda}$.
Substituting these into the expression:
$\frac{-\lambda - (-1/\lambda)}{1 + (-\lambda)(-1/\lambda)} = \frac{-\lambda + 1/\lambda}{1 + 1} = \frac{\frac{1-\lambda^2}{\lambda}}{2} = \frac{1-\lambda^2}{2\lambda}$.

(D) Given,$\sin \theta + \cos \theta = p$ $(i)$ and $\sin^3 \theta + \cos^3 \theta = q$ (ii).
Using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we have:
$(\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) = q$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,this becomes $p(1 - \sin \theta \cos \theta) = q$.
Thus,$1 - \sin \theta \cos \theta = \frac{q}{p}$,which implies $\sin \theta \cos \theta = 1 - \frac{q}{p}$ (iii).
Squaring equation $(i)$: $(\sin \theta + \cos \theta)^2 = p^2$.
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = p^2$.
$1 + 2 \sin \theta \cos \theta = p^2$.
Substituting (iii) into this: $1 + 2(1 - \frac{q}{p}) = p^2$.
$1 + 2 - \frac{2q}{p} = p^2$.
$3 - \frac{2q}{p} = p^2$.
Multiply by $p$: $3p - 2q = p^3$.
Rearranging gives $p^3 - 3p = -2q$,or $p(p^2 - 3) = -2q$.

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$
$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply and divide by $2$:
$= 2 \left( \frac{\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
$= 2 \left( \frac{\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
$= 2 \left( \frac{\sin(60^{\circ} - 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
$= 2 \left( \frac{\sin 40^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
$= 2 \left( \frac{\sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} \right)$
$= 2 \times 2 = 4$