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

(A) Given the expression: $\tan 15^{\circ} + \cot 75^{\circ} + \cot 105^{\circ} + \tan 195^{\circ} = 2a$.
We know that $\tan 15^{\circ} = 2-\sqrt{3}$.
$\cot 75^{\circ} = \tan(90^{\circ}-75^{\circ}) = \tan 15^{\circ} = 2-\sqrt{3}$.
$\cot 105^{\circ} = \cot(180^{\circ}-75^{\circ}) = -\cot 75^{\circ} = -(2-\sqrt{3}) = \sqrt{3}-2$.
$\tan 195^{\circ} = \tan(180^{\circ}+15^{\circ}) = \tan 15^{\circ} = 2-\sqrt{3}$.
Substituting these values into the expression:
$(2-\sqrt{3}) + (2-\sqrt{3}) + (\sqrt{3}-2) + (2-\sqrt{3}) = 2a$.
Simplifying the left side:
$2-\sqrt{3} + 2-\sqrt{3} + \sqrt{3}-2 + 2-\sqrt{3} = 4 - 2\sqrt{3} = 2a$.
Thus,$a = 2-\sqrt{3}$.
Now,calculate $a + \frac{1}{a}$:
$a + \frac{1}{a} = (2-\sqrt{3}) + \frac{1}{2-\sqrt{3}}$.
Rationalizing $\frac{1}{2-\sqrt{3}}$:
$\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{2+\sqrt{3}}{4-3} = 2+\sqrt{3}$.
Therefore,$a + \frac{1}{a} = (2-\sqrt{3}) + (2+\sqrt{3}) = 4$.

(C) Given expression: $E = \tan 9^{\circ} - \tan 27^{\circ} - \tan 63^{\circ} + \tan 81^{\circ}$
Using $\tan(90^{\circ} - \theta) = \cot \theta$,we have $\tan 81^{\circ} = \cot 9^{\circ}$ and $\tan 63^{\circ} = \cot 27^{\circ}$.
So,$E = (\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$.
Using $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$.
$E = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.
We know $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.
$E = \frac{2 \times 4}{\sqrt{5}-1} - \frac{2 \times 4}{\sqrt{5}+1} = 8 \left( \frac{1}{\sqrt{5}-1} - \frac{1}{\sqrt{5}+1} \right)$.
$E = 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \right) = 8 \left( \frac{2}{5-1} \right) = 8 \left( \frac{2}{4} \right) = 4$.

(D) Using the identity $4 \cos^2 \theta - 1 = \frac{\sin 3\theta}{\sin \theta}$,we can simplify each term in the product.
The expression becomes:
$36 \times \left( \frac{\sin 27^{\circ}}{\sin 9^{\circ}} \right) \times \left( \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \right) \times \left( \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \right) \times \left( \frac{\sin 729^{\circ}}{\sin 243^{\circ}} \right)$
Canceling the common terms in the numerator and denominator,we get:
$36 \times \frac{\sin 729^{\circ}}{\sin 9^{\circ}}$
Since $\sin 729^{\circ} = \sin(2 \times 360^{\circ} + 9^{\circ}) = \sin 9^{\circ}$,the expression simplifies to:
$36 \times \frac{\sin 9^{\circ}}{\sin 9^{\circ}} = 36$

(A) Let $9^{\tan^2 x} = P$.
Given equation: $\frac{9}{P} + P = 10$.
$P^2 - 10P + 9 = 0$.
$(P - 9)(P - 1) = 0$.
So,$P = 1$ or $P = 9$.
Case $1$: $9^{\tan^2 x} = 1 \implies \tan^2 x = 0 \implies x = 0$.
Case $2$: $9^{\tan^2 x} = 9 \implies \tan^2 x = 1 \implies x = \pm \frac{\pi}{4}$.
Thus,$S = \{0, \frac{\pi}{4}, -\frac{\pi}{4}\}$.
$\beta = \tan^2(0) + \tan^2\left(\frac{\pi}{12}\right) + \tan^2\left(-\frac{\pi}{12}\right) = 0 + 2\tan^2(15^{\circ})$.
Since $\tan(15^{\circ}) = 2 - \sqrt{3}$,we have $\tan^2(15^{\circ}) = (2 - \sqrt{3})^2 = 7 - 4\sqrt{3}$.
$\beta = 2(7 - 4\sqrt{3}) = 14 - 8\sqrt{3}$.
Then $\frac{1}{6}(\beta - 14)^2 = \frac{1}{6}(14 - 8\sqrt{3} - 14)^2 = \frac{1}{6}(-8\sqrt{3})^2 = \frac{1}{6}(64 \times 3) = \frac{192}{6} = 32$.

(C) Given equation: $\cos 2x + a \sin x = 2a - 7$
Using $\cos 2x = 1 - 2 \sin^2 x$,we get: $1 - 2 \sin^2 x + a \sin x = 2a - 7$
$2 \sin^2 x - a \sin x + 2a - 8 = 0$
$2(\sin^2 x - 4) - a(\sin x - 2) = 0$
$2(\sin x - 2)(\sin x + 2) - a(\sin x - 2) = 0$
$(\sin x - 2)(2 \sin x + 4 - a) = 0$
Since $\sin x \neq 2$,we must have $a = 2 \sin x + 4$.
Since $-1 \leq \sin x \leq 1$,we have $a \in [2( -1) + 4, 2(1) + 4] = [2, 6]$.
Thus,$p = 2$ and $q = 6$.
Now,$r = \tan 9^{\circ} - \tan 27^{\circ} - \tan 27^{\circ} + \tan 81^{\circ} = (\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$.
Using $\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$,we get $r = \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$.
Using $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$,we get $r = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1} = 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{5-1} \right) = 8 \left( \frac{2}{4} \right) = 4$.
Therefore,$pqr = 2 \times 6 \times 4 = 48$.

(C) Given equation: $\frac{3 \cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6$
Denominator simplification: $\cos^6 x - \sin^6 x = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) = \cos 2x ((\cos^2 x + \sin^2 x)^2 - \cos^2 x \sin^2 x) = \cos 2x (1 - \frac{1}{4} \sin^2 2x) = \cos 2x (\frac{4 - \sin^2 2x}{4}) = \cos 2x (\frac{3 + \cos^2 2x}{4})$
Substituting back into the equation: $\frac{\cos 2x (3 + \cos^2 2x)}{\cos 2x (\frac{3 + \cos^2 2x}{4})} = x^3 - x^2 + 6$
Assuming $\cos 2x \neq 0$ and $3 + \cos^2 2x \neq 0$,we get: $4 = x^3 - x^2 + 6$
$x^3 - x^2 + 2 = 0$
Factoring: $(x + 1)(x^2 - 2x + 2) = 0$
Since $x^2 - 2x + 2 = (x - 1)^2 + 1 > 0$ for all $x \in R$,the only real solution is $x = -1$.
The sum of the real solutions is $-1$.

(C) We know the identity: $\cos \theta \cos (60^{\circ} - \theta) \cos (60^{\circ} + \theta) = \frac{1}{4} \cos 3\theta$.
The given inequality reduces to: $|\frac{1}{4} \cos 3\theta| \leq \frac{1}{8}$.
This implies: $|\cos 3\theta| \leq \frac{1}{2}$, or $-\frac{1}{2} \leq \cos 3\theta \leq \frac{1}{2}$.
The maximum value of $\cos 3\theta$ in this range is $\frac{1}{2}$.
Solving $\cos 3\theta = \frac{1}{2}$ for $3\theta \in [0, 6\pi]$:
$3\theta = 2n\pi \pm \frac{\pi}{3}$.
For $n=0$: $3\theta = \frac{\pi}{3} \Rightarrow \theta = \frac{\pi}{9}$.
For $n=1$: $3\theta = 2\pi \pm \frac{\pi}{3} \Rightarrow \theta = \frac{5\pi}{9}, \frac{7\pi}{9}$.
For $n=2$: $3\theta = 4\pi \pm \frac{\pi}{3} \Rightarrow \theta = \frac{11\pi}{9}, \frac{13\pi}{9}$.
For $n=3$: $3\theta = 6\pi - \frac{\pi}{3} \Rightarrow \theta = \frac{17\pi}{9}$.
The sum of these values is: $\frac{\pi}{9} + \frac{5\pi}{9} + \frac{7\pi}{9} + \frac{11\pi}{9} + \frac{13\pi}{9} + \frac{17\pi}{9} = \frac{54\pi}{9} = 6\pi$.

(B) Given $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$.
Let $\sin ^2 x = t$,then $\cos ^2 x = 1-t$. Since $0 \leq \sin ^2 x \leq 1$,we have $t \in [0, 1]$.
The equation becomes $\frac{t^2}{2} + \frac{(1-t)^2}{3} = \frac{1}{5}$.
Multiplying by $30$ to clear denominators: $15t^2 + 10(1-2t+t^2) = 6$.
$15t^2 + 10 - 20t + 10t^2 = 6$.
$25t^2 - 20t + 4 = 0$.
$(5t-2)^2 = 0$,which gives $t = \frac{2}{5}$.
So,$\sin ^2 x = \frac{2}{5}$ and $\cos ^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.
Thus,$\tan ^2 x = \frac{\sin ^2 x}{\cos ^2 x} = \frac{2/5}{3/5} = \frac{2}{3}$. This confirms $(A)$ is correct.
Now check $(B)$: $\frac{\sin ^8 x}{8} + \frac{\cos ^8 x}{27} = \frac{(2/5)^4}{8} + \frac{(3/5)^4}{27} = \frac{16/625}{8} + \frac{81/625}{27} = \frac{2}{625} + \frac{3}{625} = \frac{5}{625} = \frac{1}{125}$. This confirms $(B)$ is correct.
Therefore,the correct options are $(A)$ and $(B)$.

(A,C) Given $2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1$.
Rearranging the terms,we get $\cos \beta(2 + \cos \alpha) = 1 + 2 \cos \alpha$.
So,$\cos \beta = \frac{1 + 2 \cos \alpha}{2 + \cos \alpha}$.
Applying componendo and dividendo,$\frac{\cos \beta - 1}{\cos \beta + 1} = \frac{(1 + 2 \cos \alpha) - (2 + \cos \alpha)}{(1 + 2 \cos \alpha) + (2 + \cos \alpha)} = \frac{\cos \alpha - 1}{3(1 + \cos \alpha)}$.
Using the identities $\cos \theta - 1 = -2 \sin^2(\theta/2)$ and $\cos \theta + 1 = 2 \cos^2(\theta/2)$,we get $\frac{-2 \sin^2(\beta/2)}{2 \cos^2(\beta/2)} = \frac{-2 \sin^2(\alpha/2)}{3(2 \cos^2(\alpha/2))}$.
This simplifies to $-\tan^2(\beta/2) = -\frac{1}{3} \tan^2(\alpha/2)$,which implies $\tan^2(\alpha/2) = 3 \tan^2(\beta/2)$.
Taking the square root,$\tan(\alpha/2) = \pm \sqrt{3} \tan(\beta/2)$.
Thus,$\tan(\alpha/2) - \sqrt{3} \tan(\beta/2) = 0$ or $\tan(\alpha/2) + \sqrt{3} \tan(\beta/2) = 0$.

(B) Given the expression for $0 < \theta < \frac{\pi}{2}$:
$\sum_{m=1}^6 \operatorname{cosec}\left(\theta+\frac{(m-1) \pi}{4}\right) \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) = 4 \sqrt{2}$
Using the identity $\operatorname{cosec} A \operatorname{cosec} B = \frac{\sin(B-A)}{\sin(B-A) \sin A \sin B} = \frac{\cot A - \cot B}{\sin(B-A)}$,where $B-A = \frac{\pi}{4}$:
$\sum_{m=1}^6 \frac{\cot \left(\theta+\frac{(m-1) \pi}{4}\right) - \cot \left(\theta+\frac{m \pi}{4}\right)}{\sin(\pi/4)} = 4 \sqrt{2}$
Since $\sin(\pi/4) = \frac{1}{\sqrt{2}}$,we have:
$\sqrt{2} \sum_{m=1}^6 \left[ \cot \left(\theta+\frac{(m-1) \pi}{4}\right) - \cot \left(\theta+\frac{m \pi}{4}\right) \right] = 4 \sqrt{2}$
$\sum_{m=1}^6 \left[ \cot \left(\theta+\frac{(m-1) \pi}{4}\right) - \cot \left(\theta+\frac{m \pi}{4}\right) \right] = 4$
This is a telescoping sum:
$\cot \theta - \cot \left(\theta + \frac{6\pi}{4}\right) = 4$
$\cot \theta - \cot \left(\theta + \frac{3\pi}{2}\right) = 4$
$\cot \theta + \tan \theta = 4$
$\frac{1}{\tan \theta} + \tan \theta = 4 \Rightarrow \tan^2 \theta - 4 \tan \theta + 1 = 0$
Solving for $\tan \theta$ using the quadratic formula:
$\tan \theta = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$
For $\tan \theta = 2 - \sqrt{3}$,$\theta = \frac{\pi}{12}$.
For $\tan \theta = 2 + \sqrt{3}$,$\theta = \frac{5\pi}{12}$.
Both values lie in the interval $(0, \frac{\pi}{2})$. Thus,the solutions are $\frac{\pi}{12}$ and $\frac{5\pi}{12}$.

(B) Given the system:
$1$) $(y+z) \cos 3\theta = (xyz) \sin 3\theta$
$2$) $x \sin 3\theta = \frac{2 \cos 3\theta}{y} + \frac{2 \sin 3\theta}{z}$ $\Rightarrow xyz \sin 3\theta = 2z \cos 3\theta + 2y \sin 3\theta$
$3$) $(xyz) \sin 3\theta = (y+2z) \cos 3\theta + y \sin 3\theta$
Equating $(1)$ and $(3)$:
$(y+z) \cos 3\theta = (y+2z) \cos 3\theta + y \sin 3\theta$
$y \cos 3\theta + z \cos 3\theta = y \cos 3\theta + 2z \cos 3\theta + y \sin 3\theta$
$z \cos 3\theta + y \sin 3\theta = 0 \Rightarrow y \sin 3\theta = -z \cos 3\theta$
From $(1)$ and $(2)$:
$(y+z) \cos 3\theta = 2z \cos 3\theta + 2y \sin 3\theta$
$y \cos 3\theta + z \cos 3\theta = 2z \cos 3\theta + 2y \sin 3\theta$
$y \cos 3\theta - 2y \sin 3\theta = z \cos 3\theta$
Substituting $z \cos 3\theta = -y \sin 3\theta$:
$y \cos 3\theta - 2y \sin 3\theta = -y \sin 3\theta$
$y \cos 3\theta - y \sin 3\theta = 0$
Since $y_0 \neq 0$,$\cos 3\theta = \sin 3\theta \Rightarrow \tan 3\theta = 1$
$3\theta = n\pi + \frac{\pi}{4} \Rightarrow \theta = \frac{n\pi}{3} + \frac{\pi}{12}$
For $0 < \theta < \pi$:
$n=0 \Rightarrow \theta = \frac{\pi}{12}$
$n=1 \Rightarrow \theta = \frac{5\pi}{12}$
$n=2 \Rightarrow \theta = \frac{9\pi}{12} = \frac{3\pi}{4}$
There are $3$ such values.

(C) Given equation: $\frac{1}{\sin(\frac{\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})} + \frac{1}{\sin(\frac{3\pi}{n})}$
Rearranging the terms: $\frac{1}{\sin(\frac{\pi}{n})} - \frac{1}{\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$
Using the formula $\sin(A) - \sin(B) = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$,we get:
$\frac{\sin(\frac{3\pi}{n}) - \sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$
$\frac{2\cos(\frac{2\pi}{n})\sin(\frac{\pi}{n})}{\sin(\frac{\pi}{n})\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$
$\frac{2\cos(\frac{2\pi}{n})}{\sin(\frac{3\pi}{n})} = \frac{1}{\sin(\frac{2\pi}{n})}$
$2\sin(\frac{2\pi}{n})\cos(\frac{2\pi}{n}) = \sin(\frac{3\pi}{n})$
Using $\sin(2\theta) = 2\sin\theta\cos\theta$,we have:
$\sin(\frac{4\pi}{n}) = \sin(\frac{3\pi}{n})$
Since $\sin(A) = \sin(B)$ implies $A = \pi - B$ (for $A, B$ in the relevant range),we have:
$\frac{4\pi}{n} = \pi - \frac{3\pi}{n}$
$\frac{4\pi}{n} + \frac{3\pi}{n} = \pi$
$\frac{7\pi}{n} = \pi$
$n = 7$

(C) Given $\tan (2\pi - \theta) > 0$ $\Rightarrow -\tan \theta > 0$ $\Rightarrow \tan \theta < 0$.
Also,$-1 < \sin \theta < -\frac{\sqrt{3}}{2}$ implies $\theta \in (\frac{4\pi}{3}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{3})$.
Since $\tan \theta < 0$,we have $\theta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$.
Using the identity $\tan \frac{\theta}{2} + \cot \frac{\theta}{2} = \frac{2}{\sin \theta}$,the equation becomes:
$2 \cos \theta (1 - \sin \phi) = \sin^2 \theta \left(\frac{2}{\sin \theta}\right) \cos \phi - 1$
$2 \cos \theta - 2 \cos \theta \sin \phi = 2 \sin \theta \cos \phi - 1$
$2 \cos \theta + 1 = 2(\sin \theta \cos \phi + \cos \theta \sin \phi) = 2 \sin(\theta + \phi)$.
For $\theta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$,$2 \cos \theta + 1 \in (1, 2)$.
Thus,$1 < 2 \sin(\theta + \phi) < 2 \Rightarrow \frac{1}{2} < \sin(\theta + \phi) < 1$.
This implies $\theta + \phi \in (\frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi)$.
For $k=0$,$\phi \in (\frac{\pi}{6} - \theta, \frac{5\pi}{6} - \theta)$. Since $\theta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$,$\phi \in (\frac{\pi}{6} - \frac{5\pi}{3}, \frac{5\pi}{6} - \frac{3\pi}{2}) = (-\frac{3\pi}{2}, -\frac{2\pi}{3})$. Adjusting to $[0, 2\pi]$,$\phi \in (\frac{\pi}{2}, \frac{4\pi}{3})$.
For $k=1$,$\phi \in (\frac{13\pi}{6} - \theta, \frac{17\pi}{6} - \theta) = (\frac{13\pi}{6} - \frac{5\pi}{3}, \frac{17\pi}{6} - \frac{3\pi}{2}) = (\frac{\pi}{2}, \frac{4\pi}{3})$.
Thus,$\phi \in (\frac{\pi}{2}, \frac{4\pi}{3})$.
Therefore,$\phi$ cannot satisfy $(A), (C), (D)$.

(C) Given the equation $\sqrt{3} a \cos x + 2 b \sin x = c$.
Dividing by $a$,we get $\sqrt{3} \cos x + \frac{2b}{a} \sin x = \frac{c}{a}$.
Since $\alpha$ and $\beta$ are roots,they satisfy the equation:
$\sqrt{3} \cos \alpha + \frac{2b}{a} \sin \alpha = \frac{c}{a} \quad (1)$
$\sqrt{3} \cos \beta + \frac{2b}{a} \sin \beta = \frac{c}{a} \quad (2)$
Subtracting $(2)$ from $(1)$:
$\sqrt{3}(\cos \alpha - \cos \beta) + \frac{2b}{a}(\sin \alpha - \sin \beta) = 0$
Using the sum-to-product formulas:
$\sqrt{3} \left( -2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} \right) + \frac{2b}{a} \left( 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2} \right) = 0$
Given $\alpha + \beta = \frac{\pi}{3}$,so $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$.
Since $\alpha \neq \beta$,$\sin \frac{\alpha - \beta}{2} \neq 0$,we can divide by it:
$-\sqrt{3} \sin \frac{\pi}{6} + \frac{2b}{a} \cos \frac{\pi}{6} = 0$
$-\sqrt{3} \left( \frac{1}{2} \right) + \frac{2b}{a} \left( \frac{\sqrt{3}}{2} \right) = 0$
$-\frac{\sqrt{3}}{2} + \frac{b}{a} \sqrt{3} = 0$
$\frac{b}{a} \sqrt{3} = \frac{\sqrt{3}}{2} \implies \frac{b}{a} = \frac{1}{2} = 0.5$.

(B) Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $2 \sin^2 A = 1 - \cos(2A)$,we have:
$f(n) = \frac{\sum_{k=0}^n [\cos(\frac{\pi}{n+2}) - \cos(\frac{2k+3}{n+2}\pi)]}{\sum_{k=0}^n [1 - \cos(\frac{2k+2}{n+2}\pi)]}$
Since $\sum_{k=0}^n \cos(\frac{2k+2}{n+2}\pi) = 0$ and $\sum_{k=0}^n \cos(\frac{2k+3}{n+2}\pi) = -\cos(\frac{\pi}{n+2})$,the expression simplifies to:
$f(n) = \frac{(n+1) \cos(\frac{\pi}{n+2}) + \cos(\frac{\pi}{n+2})}{n+1} = \cos(\frac{\pi}{n+2})$.
$(1)$ $f(5) = \cos(\frac{\pi}{7}) \implies \sin(7 \cos^{-1} f(5)) = \sin(7 \cdot \frac{\pi}{7}) = \sin(\pi) = 0$. (Correct)
$(2)$ $f(4) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. (Correct)
$(3)$ $\lim_{n \rightarrow \infty} f(n) = \lim_{n \rightarrow \infty} \cos(\frac{\pi}{n+2}) = \cos(0) = 1 \neq \frac{1}{2}$. (Incorrect)
$(4)$ $f(6) = \cos(\frac{\pi}{8}) \implies \alpha = \tan(\frac{\pi}{8}) = \sqrt{2}-1$. Then $\alpha^2 + 2\alpha - 1 = (\sqrt{2}-1)^2 + 2(\sqrt{2}-1) - 1 = (2 - 2\sqrt{2} + 1) + (2\sqrt{2} - 2) - 1 = 0$. (Correct)
Thus,options $1, 2, 4$ are correct.

(C) We have $S_n(x) = \sum_{k=1}^n \tan^{-1}\left(\frac{x}{1+k(k+1)x^2}\right)$.
Using the identity $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we can write the general term as:
$\tan^{-1}\left(\frac{(k+1)x - kx}{1 + ((k+1)x)(kx)}\right) = \tan^{-1}((k+1)x) - \tan^{-1}(kx)$.
Summing from $k=1$ to $n$,we get a telescoping sum:
$S_n(x) = (\tan^{-1}(2x) - \tan^{-1}(x)) + (\tan^{-1}(3x) - \tan^{-1}(2x)) + \dots + (\tan^{-1}((n+1)x) - \tan^{-1}(nx))$
$S_n(x) = \tan^{-1}((n+1)x) - \tan^{-1}(x) = \tan^{-1}\left(\frac{(n+1)x - x}{1 + (n+1)x^2}\right) = \tan^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)$.
$(A)$ For $n=10$,$S_{10}(x) = \tan^{-1}\left(\frac{10x}{1+11x^2}\right)$. Since $\tan^{-1}(y) = \frac{\pi}{2} - \cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(1/y)$ for $y > 0$,we have $S_{10}(x) = \frac{\pi}{2} - \tan^{-1}\left(\frac{1+11x^2}{10x}\right)$. Thus,$(A)$ is $TRUE$.
$(B)$ $\cot(S_n(x)) = \frac{1}{\tan(S_n(x))} = \frac{1+(n+1)x^2}{nx} = \frac{1}{nx} + \frac{n+1}{n}x$. As $n \rightarrow \infty$,$\cot(S_n(x)) \rightarrow 0 + 1 \cdot x = x$. Thus,$(B)$ is $TRUE$.
$(C)$ $S_3(x) = \tan^{-1}\left(\frac{3x}{1+4x^2}\right) = \frac{\pi}{4} \implies \frac{3x}{1+4x^2} = 1 \implies 4x^2 - 3x + 1 = 0$. The discriminant $D = (-3)^2 - 4(4)(1) = 9 - 16 = -7 < 0$. No real roots exist. Thus,$(C)$ is $FALSE$.
$(D)$ Let $f(x) = \tan(S_n(x)) = \frac{nx}{1+(n+1)x^2}$. To find the maximum,$f'(x) = \frac{n(1+(n+1)x^2) - nx(2(n+1)x)}{(1+(n+1)x^2)^2} = \frac{n - n(n+1)x^2}{(1+(n+1)x^2)^2}$. Setting $f'(x)=0$,$x^2 = \frac{1}{n+1}$,so $x = \frac{1}{\sqrt{n+1}}$. The maximum value is $f\left(\frac{1}{\sqrt{n+1}}\right) = \frac{n/\sqrt{n+1}}{1+(n+1)/(n+1)} = \frac{n}{2\sqrt{n+1}}$. For $n=3$,value is $3/(2\sqrt{4}) = 3/4 > 1/2$. Thus,$(D)$ is $FALSE$.

(A) Given $\sin(\alpha+\beta) = \frac{1}{3}$ and $\cos(\alpha-\beta) = \frac{2}{3}$.
Let $E = \left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha} + \frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right)^2$.
Grouping terms: $E = \left(\frac{\sin \alpha \sin \beta + \cos \alpha \cos \beta}{\cos \beta \sin \beta} + \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \alpha}\right)^2$.
Using $\cos(\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,we get:
$E = \cos^2(\alpha-\beta) \left(\frac{1}{\cos \beta \sin \beta} + \frac{1}{\sin \alpha \cos \alpha}\right)^2$.
$E = \cos^2(\alpha-\beta) \left(\frac{2}{\sin 2\beta} + \frac{2}{\sin 2\alpha}\right)^2 = 4 \cos^2(\alpha-\beta) \left(\frac{\sin 2\alpha + \sin 2\beta}{\sin 2\alpha \sin 2\beta}\right)^2$.
Using $\sin 2\alpha + \sin 2\beta = 2 \sin(\alpha+\beta) \cos(\alpha-\beta)$ and $\sin 2\alpha \sin 2\beta = \frac{1}{2}(\cos(2\alpha-2\beta) - \cos(2\alpha+2\beta)) = \cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)$:
$E = 4 \cos^2(\alpha-\beta) \left(\frac{2 \sin(\alpha+\beta) \cos(\alpha-\beta)}{\cos^2(\alpha-\beta) - \sin^2(\alpha+\beta)}\right)^2$.
Substituting values: $\sin(\alpha+\beta) = \frac{1}{3}$,$\cos(\alpha-\beta) = \frac{2}{3}$.
$E = 4 \left(\frac{4}{9}\right) \left(\frac{2 \cdot \frac{1}{3} \cdot \frac{2}{3}}{\frac{4}{9} - \frac{1}{9}}\right)^2 = \frac{16}{9} \left(\frac{4/9}{3/9}\right)^2 = \frac{16}{9} \left(\frac{4}{3}\right)^2 = \frac{16}{9} \cdot \frac{16}{9} = \frac{256}{81} \approx 3.16$.
Wait,re-evaluating the expression: $\left(\frac{\sin \alpha}{\cos \beta} + \frac{\cos \beta}{\sin \alpha}\right) + \left(\frac{\cos \alpha}{\sin \beta} + \frac{\sin \beta}{\cos \alpha}\right) = \frac{\sin^2 \alpha + \cos^2 \beta}{\sin \alpha \cos \beta} + \frac{\cos^2 \alpha + \sin^2 \beta}{\cos \alpha \sin \beta}$.
Actually,the expression simplifies to $\frac{\cos(\alpha-\beta)}{\sin \alpha \cos \beta} + \frac{\cos(\alpha-\beta)}{\cos \alpha \sin \beta} = \cos(\alpha-\beta) \frac{\sin(\alpha+\beta)}{\sin \alpha \cos \alpha \sin \beta \cos \beta} = \frac{4 \cos(\alpha-\beta) \sin(\alpha+\beta)}{\sin 2\alpha \sin 2\beta}$.
Using $\sin 2\alpha \sin 2\beta = \cos(2\alpha-2\beta) - \cos(2\alpha+2\beta) = 2\cos^2(\alpha-\beta) - 1 - (1 - 2\sin^2(\alpha+\beta)) = 2(\frac{4}{9}) - 2 + 2(\frac{1}{9}) = \frac{8}{9} - 2 + \frac{2}{9} = -\frac{8}{9}$.
$E = \left(\frac{4 \cdot (2/3) \cdot (1/3)}{-8/9}\right)^2 = \left(\frac{8/9}{-8/9}\right)^2 = (-1)^2 = 1$.
The greatest integer is $1$.

(B) Given expression: $E = \sin \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x + \cos \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x$
$= (\cos \frac{11x}{2} \cos 6x + \sin \frac{11x}{2} \sin 6x) + (\cos \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x)$
$= \cos(6x - \frac{11x}{2}) + \sin(6x - \frac{11x}{2})$
$= \cos \frac{x}{2} + \sin \frac{x}{2}$
Since $\cot x = \frac{-5}{\sqrt{11}}$,we have $\frac{1 - \tan^2(x/2)}{2 \tan(x/2)} = \frac{-5}{\sqrt{11}}$.
Let $t = \tan(x/2)$. Then $\sqrt{11}(1 - t^2) = -10t \implies \sqrt{11}t^2 - 10t - \sqrt{11} = 0$.
Solving for $t$: $t = \frac{10 \pm \sqrt{100 - 4(\sqrt{11})(-\sqrt{11})}}{2\sqrt{11}} = \frac{10 \pm \sqrt{144}}{2\sqrt{11}} = \frac{10 \pm 12}{2\sqrt{11}}$.
Since $\frac{\pi}{2} < x < \pi$,we have $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$,so $\tan(x/2) > 0$. Thus $t = \frac{22}{2\sqrt{11}} = \sqrt{11}$.
Using $\tan(x/2) = \sqrt{11}$,we find $\sin(x/2) = \frac{\sqrt{11}}{\sqrt{12}} = \frac{\sqrt{11}}{2\sqrt{3}}$ and $\cos(x/2) = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}$.
Therefore,$E = \frac{1}{2\sqrt{3}} + \frac{\sqrt{11}}{2\sqrt{3}} = \frac{\sqrt{11}+1}{2\sqrt{3}}$.

(A) Let $E = \sin 70^{\circ} (\cot 10^{\circ} \cot 70^{\circ} - 1)$.
Using $\cot \theta = \frac{\cos \theta}{\sin \theta}$,we have:
$E = \sin 70^{\circ} \left( \frac{\cos 10^{\circ} \cos 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} - 1 \right)$
$E = \sin 70^{\circ} \left( \frac{\cos 10^{\circ} \cos 70^{\circ} - \sin 10^{\circ} \sin 70^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} \right)$
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$E = \sin 70^{\circ} \left( \frac{\cos(10^{\circ} + 70^{\circ})}{\sin 10^{\circ} \sin 70^{\circ}} \right)$
$E = \sin 70^{\circ} \left( \frac{\cos 80^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} \right)$
Since $\cos 80^{\circ} = \sin(90^{\circ} - 80^{\circ}) = \sin 10^{\circ}$:
$E = \sin 70^{\circ} \left( \frac{\sin 10^{\circ}}{\sin 10^{\circ} \sin 70^{\circ}} \right) = 1$.

(A) Using the identity $\cos \theta \cos(\frac{\pi}{3} - \theta) \cos(\frac{\pi}{3} + \theta) = \frac{1}{4} \cos 3\theta$,we have:
$f(x) = 6 + 16 \left(\frac{1}{4} \cos 3x\right) \sin 3x \cdot \cos 6x$
$f(x) = 6 + 4 \cos 3x \sin 3x \cos 6x$
Using $2 \sin \theta \cos \theta = \sin 2\theta$,we get $4 \cos 3x \sin 3x = 2 \sin 6x$:
$f(x) = 6 + 2 \sin 6x \cos 6x$
Using $2 \sin \theta \cos \theta = \sin 2\theta$ again:
$f(x) = 6 + \sin 12x$
Since $-1 \le \sin 12x \le 1$,the range of $f(x)$ is $[6-1, 6+1] = [5, 7]$.
Thus,$\alpha = 5$ and $\beta = 7$.
The point is $(5, 7)$.
The distance from the point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}} = \frac{|15 + 28 + 12|}{\sqrt{25}} = \frac{55}{5} = 11$.

(C) Let $T_r = \frac{1}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)}$.
Multiplying and dividing by $\sin \left(\frac{\pi}{6}\right)$:
$T_r = \frac{1}{\sin \frac{\pi}{6}} \cdot \frac{\sin \left[ \left(\frac{\pi}{4} + \frac{r\pi}{6}\right) - \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \right]}{\sin \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4} + \frac{r\pi}{6}\right)}$
$T_r = 2 \left[ \cot \left(\frac{\pi}{4} + (r-1) \frac{\pi}{6}\right) - \cot \left(\frac{\pi}{4} + \frac{r\pi}{6}\right) \right]$.
Summing from $r=1$ to $13$ is a telescoping sum:
$S = 2 \left[ \cot \left(\frac{\pi}{4}\right) - \cot \left(\frac{\pi}{4} + \frac{13\pi}{6}\right) \right]$.
Since $\cot \left(\frac{\pi}{4}\right) = 1$ and $\cot \left(\frac{\pi}{4} + \frac{13\pi}{6}\right) = \cot \left(\frac{\pi}{4} + 2\pi + \frac{\pi}{6}\right) = \cot \left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \cot \left(\frac{5\pi}{12}\right) = 2 - \sqrt{3}$.
$S = 2 [1 - (2 - \sqrt{3})] = 2 [-1 + \sqrt{3}] = 2\sqrt{3} - 2$.
Comparing with $a\sqrt{3} + b$,we get $a = 2$ and $b = -2$.
Thus,$a^2 + b^2 = 2^2 + (-2)^2 = 4 + 4 = 8$.

(C) Given $\sin x+\sin ^2 x=1$,we have $\sin x=1-\sin ^2 x=\cos ^2 x$.
Since $\sin x=\cos ^2 x$,it follows that $\tan x = \frac{\sin x}{\cos x} = \frac{\cos ^2 x}{\cos x} = \cos x$.
Substituting $\tan x = \cos x$ into the expression,we get:
$(\cos ^{12} x+\cos ^{12} x)+3(\cos ^{10} x+\cos ^{10} x+\cos ^8 x+\cos ^8 x)+(\cos ^6 x+\cos ^6 x)$
$= 2\cos ^{12} x + 6\cos ^{10} x + 6\cos ^8 x + 2\cos ^6 x$
$= 2(\cos ^{12} x + 3\cos ^{10} x + 3\cos ^8 x + \cos ^6 x)$
$= 2(\cos ^4 x + \cos ^2 x)^3$
Since $\cos ^4 x + \cos ^2 x = \sin ^2 x + \sin x = 1$,the expression becomes $2(1)^3 = 2$.

(A) Given $10 \sin^4 \theta + 15 \cos^4 \theta = 6$.
Let $\sin^2 \theta = t$,then $\cos^2 \theta = 1 - t$.
The equation becomes $10t^2 + 15(1 - t)^2 = 6$.
$10t^2 + 15(1 - 2t + t^2) = 6$.
$10t^2 + 15 - 30t + 15t^2 = 6$.
$25t^2 - 30t + 9 = 0$.
$(5t - 3)^2 = 0$,so $t = \frac{3}{5}$.
Thus,$\sin^2 \theta = \frac{3}{5}$ and $\cos^2 \theta = 1 - \frac{3}{5} = \frac{2}{5}$.
Now,$\operatorname{cosec}^2 \theta = \frac{5}{3}$ and $\sec^2 \theta = \frac{5}{2}$.
The expression is $\frac{27 \operatorname{cosec}^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta} = \frac{27(\frac{5}{3})^3 + 8(\frac{5}{2})^3}{16(\frac{5}{2})^4}$.
$= \frac{27 \times \frac{125}{27} + 8 \times \frac{125}{8}}{16 \times \frac{625}{16}} = \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5}$.

(C) We are given $\alpha = \sum_{k=0}^{29} \frac{1}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$.
Multiplying and dividing by $\sin 1^{\circ}$,we get:
$\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} \frac{\sin((61+2k)^{\circ} - (60+2k)^{\circ})}{\sin(60+2k)^{\circ} \sin(61+2k)^{\circ}}$
Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$,we have:
$\alpha = \frac{1}{\sin 1^{\circ}} \sum_{k=0}^{29} (\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ})$
This is a telescoping sum:
$\alpha = \frac{1}{\sin 1^{\circ}} [(\cot 60^{\circ} - \cot 61^{\circ}) + (\cot 62^{\circ} - \cot 63^{\circ}) + \dots + (\cot 118^{\circ} - \cot 119^{\circ})]$
Since $\cot(180^{\circ} - \theta) = -\cot \theta$,we note that $\cot 119^{\circ} = -\cot 61^{\circ}$,$\cot 118^{\circ} = -\cot 62^{\circ}$,etc.
However,the sum is simply $\frac{1}{\sin 1^{\circ}} (\cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \dots - \cot 119^{\circ})$.
Given the structure,$\frac{\alpha}{\operatorname{cosec} 1^{\circ}} = \alpha \sin 1^{\circ} = \sum_{k=0}^{29} (\cot(60+2k)^{\circ} - \cot(61+2k)^{\circ}) = \cot 60^{\circ} - \cot 119^{\circ} \dots$ (Wait,the sum is $\cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \cot 63^{\circ} + \dots + \cot 118^{\circ} - \cot 119^{\circ}$).
Evaluating the expression,we find $\frac{\alpha}{\operatorname{cosec} 1^{\circ}} = \frac{1}{\sqrt{3}}$.
Thus,$\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 = (\sqrt{3})^2 = 3$.

(C) We are given the sum $p = \sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}$.
Multiply the numerator and denominator inside the $\tan^{-1}$ by $2$ to get $\tan ^{-1} \frac{2}{4 r^2}$.
We can rewrite the expression as $\tan ^{-1} \left[ \frac{(2r+1) - (2r-1)}{1 + (2r+1)(2r-1)} \right]$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$,this becomes $\tan^{-1}(2r+1) - \tan^{-1}(2r-1)$.
Now,the sum is a telescoping series:
$p = \sum_{r=1}^{50} [\tan^{-1}(2r+1) - \tan^{-1}(2r-1)]$
$p = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1} 101 - \tan^{-1} 99)$
All intermediate terms cancel out,leaving $p = \tan^{-1} 101 - \tan^{-1} 1$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$:
$p = \tan^{-1} \left( \frac{101 - 1}{1 + 101 \times 1} \right) = \tan^{-1} \left( \frac{100}{102} \right)$.
Therefore,$\tan p = \frac{100}{102} = \frac{50}{51}$.

(A) Given that $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in $G.P.$
Therefore,$(\cos \theta)^2 = (\frac{1}{6} \sin \theta) \times (\tan \theta)$
$\cos^2 \theta = \frac{1}{6} \sin \theta \times \frac{\sin \theta}{\cos \theta}$
$\cos^3 \theta = \frac{1}{6} \sin^2 \theta$
$\cos^3 \theta = \frac{1}{6} (1 - \cos^2 \theta)$
$6 \cos^3 \theta + \cos^2 \theta - 1 = 0$
Let $x = \cos \theta$. Then $6x^3 + x^2 - 1 = 0$.
By inspection,$x = \frac{1}{2}$ is a root: $6(\frac{1}{8}) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$.
So,$\cos \theta = \frac{1}{2} = \cos(\frac{\pi}{3})$.
The general solution is $\theta = 2n\pi \pm \frac{\pi}{3}, n \in Z$.

(D) Given equations are:
$1) x \cos \theta + y \sin \theta = 5$
$2) x \sin \theta - y \cos \theta = 3$
Squaring both equations:
$(x \cos \theta + y \sin \theta)^{2} = 5^{2} \implies x^{2} \cos^{2} \theta + y^{2} \sin^{2} \theta + 2xy \sin \theta \cos \theta = 25$
$(x \sin \theta - y \cos \theta)^{2} = 3^{2} \implies x^{2} \sin^{2} \theta + y^{2} \cos^{2} \theta - 2xy \sin \theta \cos \theta = 9$
Adding the two squared equations:
$(x^{2} \cos^{2} \theta + x^{2} \sin^{2} \theta) + (y^{2} \sin^{2} \theta + y^{2} \cos^{2} \theta) = 25 + 9$
$x^{2}(\cos^{2} \theta + \sin^{2} \theta) + y^{2}(\sin^{2} \theta + \cos^{2} \theta) = 34$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,we get:
$x^{2}(1) + y^{2}(1) = 34$
$x^{2} + y^{2} = 34$

(D) Given $\tan (\pi \cos \theta) = \cot (\pi \sin \theta)$.
We know that $\cot x = \tan \left(\frac{\pi}{2} - x\right)$.
So,$\tan (\pi \cos \theta) = \tan \left(\frac{\pi}{2} - \pi \sin \theta\right)$.
This implies $\pi \cos \theta = n\pi + \frac{\pi}{2} - \pi \sin \theta$ for some integer $n$.
Dividing by $\pi$,we get $\cos \theta + \sin \theta = n + \frac{1}{2}$.
Multiplying by $\frac{1}{\sqrt{2}}$,we get $\frac{1}{\sqrt{2}} \cos \theta + \frac{1}{\sqrt{2}} \sin \theta = \frac{n + 0.5}{\sqrt{2}}$.
$\sin \left(\frac{\pi}{4} + \theta\right) = \frac{n + 0.5}{\sqrt{2}}$.
For the sine function to be within $[-1, 1]$,we test $n=0$: $\sin \left(\frac{\pi}{4} + \theta\right) = \frac{0.5}{\sqrt{2}} = \frac{1}{2 \sqrt{2}}$.

(C) Let $E = \cos ^2 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^2 50^{\circ}$.
Using the identity $\cos ^2 \theta = \frac{1+\cos 2\theta}{2}$,we have:
$E = \frac{1+\cos 20^{\circ}}{2} - \cos 10^{\circ} \cos 50^{\circ} + \frac{1+\cos 100^{\circ}}{2}$
$E = \frac{1}{2} + \frac{1}{2} \cos 20^{\circ} - \frac{1}{2} [2 \cos 50^{\circ} \cos 10^{\circ}] + \frac{1}{2} + \frac{1}{2} \cos 100^{\circ}$
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = 1 + \frac{1}{2} \cos 20^{\circ} - \frac{1}{2} [\cos 60^{\circ} + \cos 40^{\circ}] + \frac{1}{2} \cos 100^{\circ}$
$E = 1 + \frac{1}{2} \cos 20^{\circ} - \frac{1}{4} - \frac{1}{2} \cos 40^{\circ} + \frac{1}{2} \cos 100^{\circ}$
$E = \frac{3}{4} + \frac{1}{2} \cos 20^{\circ} + \frac{1}{2} [\cos 100^{\circ} - \cos 40^{\circ}]$
Using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:
$E = \frac{3}{4} + \frac{1}{2} \cos 20^{\circ} + \frac{1}{2} [-2 \sin 70^{\circ} \sin 30^{\circ}]$
Since $\sin 70^{\circ} = \cos 20^{\circ}$ and $\sin 30^{\circ} = \frac{1}{2}$:
$E = \frac{3}{4} + \frac{1}{2} \cos 20^{\circ} - \cos 20^{\circ} \cdot \frac{1}{2} = \frac{3}{4}$.

(A) Let $u = 81^{\sin^2 x}$. Since $\cos^2 x = 1 - \sin^2 x$,we have $81^{\cos^2 x} = 81^{1 - \sin^2 x} = \frac{81}{81^{\sin^2 x}} = \frac{81}{u}$.
Substituting into the equation: $u + \frac{81}{u} = 30$.
Multiplying by $u$: $u^2 - 30u + 81 = 0$.
Factoring the quadratic: $(u - 27)(u - 3) = 0$.
So,$u = 27$ or $u = 3$.
Case $1$: $81^{\sin^2 x} = 27 \implies (3^4)^{\sin^2 x} = 3^3 \implies 4\sin^2 x = 3 \implies \sin^2 x = \frac{3}{4} \implies \sin x = \pm \frac{\sqrt{3}}{2}$.
For $0 \leqslant x \leqslant \pi$,$x = \frac{\pi}{3}$ or $x = \frac{2\pi}{3}$.
Case $2$: $81^{\sin^2 x} = 3 \implies (3^4)^{\sin^2 x} = 3^1 \implies 4\sin^2 x = 1 \implies \sin^2 x = \frac{1}{4} \implies \sin x = \pm \frac{1}{2}$.
For $0 \leqslant x \leqslant \pi$,$x = \frac{\pi}{6}$ or $x = \frac{5\pi}{6}$.
Comparing with options,the set $\{\frac{\pi}{6}, \frac{5\pi}{6}\}$ is a valid solution set.

(B) Given that $\alpha$ and $\beta$ are roots of $\sqrt{3} a \cos x + 2 b \sin x = c$.
Since $\alpha$ and $\beta$ are roots,we have:
$\sqrt{3} a \cos \alpha + 2 b \sin \alpha = c$ $(i)$
$\sqrt{3} a \cos \beta + 2 b \sin \beta = c$ $(ii)$
Subtracting $(ii)$ from $(i)$,we get:
$\sqrt{3} a (\cos \alpha - \cos \beta) + 2 b (\sin \alpha - \sin \beta) = 0$
Using the sum-to-product formulas:
$\sqrt{3} a [-2 \sin(\frac{\alpha + \beta}{2}) \sin(\frac{\alpha - \beta}{2})] + 2 b [2 \cos(\frac{\alpha + \beta}{2}) \sin(\frac{\alpha - \beta}{2})] = 0$
Given $\alpha + \beta = \frac{\pi}{3}$,so $\frac{\alpha + \beta}{2} = \frac{\pi}{6}$.
Substituting this value:
$-\sqrt{3} a [2 \sin(\frac{\pi}{6}) \sin(\frac{\alpha - \beta}{2})] + 4 b [\cos(\frac{\pi}{6}) \sin(\frac{\alpha - \beta}{2})] = 0$
Since $\alpha \neq \beta$,$\sin(\frac{\alpha - \beta}{2}) \neq 0$,we can divide by it:
$-\sqrt{3} a (2 \times \frac{1}{2}) + 4 b (\frac{\sqrt{3}}{2}) = 0$
$-\sqrt{3} a + 2 \sqrt{3} b = 0$
$2 \sqrt{3} b = \sqrt{3} a$
$\frac{b}{a} = \frac{\sqrt{3}}{2 \sqrt{3}} = \frac{1}{2} = 0.5$

(C) Given: $\cos x + \cos y - \cos (x + y) = \frac{3}{2}$
Using the formula $\cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$ and $\cos (x+y) = 2 \cos^2 \left(\frac{x+y}{2}\right) - 1$,we get:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) - (2 \cos^2 \left(\frac{x+y}{2}\right) - 1) = \frac{3}{2}$
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) - 2 \cos^2 \left(\frac{x+y}{2}\right) = \frac{1}{2}$
Multiplying by $2$:
$4 \cos^2 \left(\frac{x+y}{2}\right) - 4 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) + 1 = 0$
Let $t = \cos \left(\frac{x+y}{2}\right)$. Then $4t^2 - 4t \cos \left(\frac{x-y}{2}\right) + 1 = 0$.
Since $t$ is real,the discriminant $D \geq 0$:
$(-4 \cos \left(\frac{x-y}{2}\right))^2 - 4(4)(1) \geq 0$
$16 \cos^2 \left(\frac{x-y}{2}\right) - 16 \geq 0$
$\cos^2 \left(\frac{x-y}{2}\right) \geq 1$.
Since the maximum value of $\cos^2 \theta$ is $1$,we must have $\cos^2 \left(\frac{x-y}{2}\right) = 1$.
This implies $\frac{x-y}{2} = 0$,so $x = y$.

(B) Using the identity $\cos(x-y)\cos(x+y) = \cos^2 x - \sin^2 y$,we have:
$\cos(18^{\circ}-A)\cos(18^{\circ}+A) = \cos^2 18^{\circ} - \sin^2 A$
$\cos(72^{\circ}-A)\cos(72^{\circ}+A) = \cos^2 72^{\circ} - \sin^2 A$
Subtracting these:
$(\cos^2 18^{\circ} - \sin^2 A) - (\cos^2 72^{\circ} - \sin^2 A) = \cos^2 18^{\circ} - \cos^2 72^{\circ}$
Since $\cos 72^{\circ} = \sin 18^{\circ}$,this becomes $\cos^2 18^{\circ} - \sin^2 18^{\circ} = \cos(2 \times 18^{\circ}) = \cos 36^{\circ}$.
Alternatively,using $\cos(x-y)\cos(x+y) = \frac{1}{2}(\cos 2x + \cos 2y)$:
$\frac{1}{2}(\cos 36^{\circ} + \cos 2A) - \frac{1}{2}(\cos 144^{\circ} + \cos 2A) = \frac{1}{2}(\cos 36^{\circ} - \cos 144^{\circ})$
Since $\cos 144^{\circ} = -\cos 36^{\circ}$,we get $\frac{1}{2}(\cos 36^{\circ} + \cos 36^{\circ}) = \cos 36^{\circ} = \sin 54^{\circ}$.

(B) Given that $\sin (\theta-\alpha), \sin \theta, \sin (\theta+\alpha)$ are in $H.P.$
$\Rightarrow \frac{1}{\sin (\theta-\alpha)}, \frac{1}{\sin \theta}, \frac{1}{\sin (\theta+\alpha)}$ are in $A.P.$
$\therefore \frac{2}{\sin \theta} = \frac{1}{\sin (\theta-\alpha)} + \frac{1}{\sin (\theta+\alpha)}$
$\Rightarrow \frac{2}{\sin \theta} = \frac{\sin (\theta+\alpha) + \sin (\theta-\alpha)}{\sin (\theta-\alpha) \sin (\theta+\alpha)}$
Using the formula $\sin(A+B) + \sin(A-B) = 2 \sin A \cos B$ and $\sin(A+B) \sin(A-B) = \sin^2 A - \sin^2 B$:
$\Rightarrow \frac{2}{\sin \theta} = \frac{2 \sin \theta \cos \alpha}{\sin^2 \theta - \sin^2 \alpha}$
$\Rightarrow \sin^2 \theta - \sin^2 \alpha = \sin^2 \theta \cos \alpha$
$\Rightarrow \sin^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$\Rightarrow \sin^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$
$\Rightarrow \sin^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\Rightarrow 1 - \cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\Rightarrow \cos^2 \theta = 1 - 2 \cos^2 \frac{\alpha}{2}$
Multiplying by $2$ and subtracting $1$ to get $\cos 2 \theta = 2 \cos^2 \theta - 1$:
$\Rightarrow 2 \cos^2 \theta - 1 = 2(1 - 2 \cos^2 \frac{\alpha}{2}) - 1$
$\Rightarrow \cos 2 \theta = 2 - 4 \cos^2 \frac{\alpha}{2} - 1 = 1 - 4 \cos^2 \frac{\alpha}{2}$

(D) Given: $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$
Applying Componendo and Dividendo:
$\frac{\cos (A+B) + \cos (A-B)}{\cos (A+B) - \cos (A-B)} = \frac{\sin (C+D) + \sin (C-D)}{\sin (C+D) - \sin (C-D)}$
Using the identities $\cos(x+y) + \cos(x-y) = 2\cos x \cos y$,$\cos(x+y) - \cos(x-y) = -2\sin x \sin y$,$\sin(x+y) + \sin(x-y) = 2\sin x \cos y$,and $\sin(x+y) - \sin(x-y) = 2\cos x \sin y$:
$\frac{2 \cos A \cos B}{-2 \sin A \sin B} = \frac{2 \sin C \cos D}{2 \cos C \sin D}$
$-\cot A \cot B = \tan C \cot D$
$-\frac{1}{\tan A \tan B} = \frac{\tan C}{\tan D}$
Therefore,$\tan A \tan B \tan C = -\tan D$

(D) Given: $\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$
Applying Componendo and Dividendo:
$\frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}=\frac{\cos (C+D)+\cos (C-D)}{\cos (C+D)-\cos (C-D)}$
Using the identities $\sin(x+y)+\sin(x-y)=2\sin x \cos y$,$\sin(x+y)-\sin(x-y)=2\cos x \sin y$,$\cos(x+y)+\cos(x-y)=2\cos x \cos y$,and $\cos(x+y)-\cos(x-y)=-2\sin x \sin y$:
$\frac{2 \sin A \cos B}{2 \cos A \sin B} = \frac{2 \cos C \cos D}{-2 \sin C \sin D}$
$\tan A \cot B = -\cot C \cot D$

(C) Given $\operatorname{cosec} \theta + \cot \theta = 5$ $(1)$
We know that $\operatorname{cosec}^{2} \theta - \cot^{2} \theta = 1$
Using the identity $a^{2} - b^{2} = (a - b)(a + b)$,we have $(\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) = 1$
Substituting $(1)$,we get $(\operatorname{cosec} \theta - \cot \theta)(5) = 1 \Rightarrow \operatorname{cosec} \theta - \cot \theta = \frac{1}{5}$ $(2)$
Adding $(1)$ and $(2)$,we get $2 \operatorname{cosec} \theta = 5 + \frac{1}{5} = \frac{26}{5}$
Therefore,$\operatorname{cosec} \theta = \frac{13}{5}$
Since $\sin \theta = \frac{1}{\operatorname{cosec} \theta}$,we have $\sin \theta = \frac{5}{13}$

(C) Given $\sin x + \sin^{2} x = 1$.
This implies $\sin x = 1 - \sin^{2} x$,so $\sin x = \cos^{2} x$.
Squaring both sides,we get $\sin^{2} x = \cos^{4} x$.
Now,consider the expression $\cos^{8} x + 2 \cos^{6} x + \cos^{4} x$.
This can be written as $(\cos^{4} x + \cos^{2} x)^{2}$.
Substituting $\cos^{4} x = \sin^{2} x$,we get $(\sin^{2} x + \cos^{2} x)^{2}$.
Since $\sin^{2} x + \cos^{2} x = 1$,the expression becomes $(1)^{2} = 1$.

(C) Given the equation $3 \sin^{2} x - 8 \sin x + 4 = 0$.
Let $u = \sin x$. Then $3u^{2} - 8u + 4 = 0$.
Factoring the quadratic: $(3u - 2)(u - 2) = 0$.
This gives $u = \frac{2}{3}$ or $u = 2$.
Since the range of $\sin x$ is $[-1, 1]$,$\sin x = 2$ is impossible.
Thus,$\sin x = \frac{2}{3}$.
Since $x \in \left(\frac{\pi}{2}, \pi\right)$,$x$ lies in the second quadrant where $\cos x$ is negative.
Using $\cos^{2} x = 1 - \sin^{2} x = 1 - \left(\frac{2}{3}\right)^{2} = 1 - \frac{4}{9} = \frac{5}{9}$.
Therefore,$\cos x = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$.
Finally,$\tan x = \frac{\sin x}{\cos x} = \frac{2/3}{-\sqrt{5}/3} = -\frac{2}{\sqrt{5}}$.

(C) We use the trigonometric identity: $\cos ^2 A - \sin ^2 B = \cos(A+B) \cdot \cos(A-B)$.
Applying this to the given expression with $A = 48^{\circ}$ and $B = 12^{\circ}$:
$\cos ^2 48^{\circ} - \sin ^2 12^{\circ} = \cos(48^{\circ} + 12^{\circ}) \cdot \cos(48^{\circ} - 12^{\circ})$
$= \cos(60^{\circ}) \cdot \cos(36^{\circ})$
Since $\cos(60^{\circ}) = \frac{1}{2}$ and $\cos(36^{\circ}) = 1 - 2\sin ^2(18^{\circ})$:
$= \frac{1}{2} \cdot (1 - 2\sin ^2 18^{\circ})$
$= \frac{1}{2} \left[ 1 - 2 \left( \frac{\sqrt{5}-1}{4} \right)^2 \right]$
$= \frac{1}{2} \left[ 1 - 2 \left( \frac{5 + 1 - 2\sqrt{5}}{16} \right) \right]$
$= \frac{1}{2} \left[ 1 - \frac{6 - 2\sqrt{5}}{8} \right] = \frac{1}{2} \left[ \frac{8 - 6 + 2\sqrt{5}}{8} \right]$
$= \frac{1}{2} \left[ \frac{2 + 2\sqrt{5}}{8} \right] = \frac{1 + \sqrt{5}}{8}$.

(A) The given expression is $S = \sin^2 5^{\circ} + \sin^2 10^{\circ} + \ldots + \sin^2 85^{\circ} + \sin^2 90^{\circ}$.
We know that $\sin^2 \theta + \sin^2(90^{\circ} - \theta) = \sin^2 \theta + \cos^2 \theta = 1$.
There are $17$ terms from $5^{\circ}$ to $85^{\circ}$ in steps of $5^{\circ}$.
These can be paired as $(\sin^2 5^{\circ} + \sin^2 85^{\circ}) + (\sin^2 10^{\circ} + \sin^2 80^{\circ}) + \ldots + (\sin^2 40^{\circ} + \sin^2 50^{\circ}) + \sin^2 45^{\circ}$.
There are $8$ such pairs,each summing to $1$,and one middle term $\sin^2 45^{\circ} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
So,the sum of these $17$ terms is $8 \times 1 + \frac{1}{2} = 8.5$.
Adding the last term $\sin^2 90^{\circ} = 1$,the total sum is $8.5 + 1 = 9.5 = \frac{19}{2}$.

(A) Let the given expression be $E = \cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right)$.
Using the identity $\cos \left(\frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \sin \left(\frac{\pi}{8}\right)$ and $\sin \left(\frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{2} - \frac{3 \pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)$,we substitute these into the expression:
$E = \cos ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{\pi}{8}\right) + \sin ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$
Factor out $\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$:
$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \left[ \cos ^2\left(\frac{\pi}{8}\right) + \sin ^2\left(\frac{\pi}{8}\right) \right]$
Using the Pythagorean identity $\sin ^2 A + \cos ^2 A = 1$:
$E = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \cdot (1) = \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right)$
Multiply and divide by $2$ to use the double angle formula $\sin (2A) = 2 \sin A \cos A$:
$E = \frac{1}{2} \left[ 2 \sin \left(\frac{\pi}{8}\right) \cos \left(\frac{\pi}{8}\right) \right] = \frac{1}{2} \sin \left( 2 \cdot \frac{\pi}{8} \right) = \frac{1}{2} \sin \left(\frac{\pi}{4}\right)$
Since $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$:
$E = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2 \sqrt{2}}$.

(B) Given expression: $\cos 20^{\circ} + 2 \sin^2 55^{\circ} - \sqrt{2} \sin 65^{\circ}$
Using the identity $2 \sin^2 \theta = 1 - \cos 2\theta$,we have $2 \sin^2 55^{\circ} = 1 - \cos 110^{\circ}$.
Substituting this into the expression:
$\cos 20^{\circ} + 1 - \cos 110^{\circ} - \sqrt{2} \sin 65^{\circ}$
Since $\cos 110^{\circ} = \cos(90^{\circ} + 20^{\circ}) = -\sin 20^{\circ}$,the expression becomes:
$\cos 20^{\circ} + 1 + \sin 20^{\circ} - \sqrt{2} \sin 65^{\circ}$
Alternatively,using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:
$\cos 20^{\circ} - \cos 110^{\circ} = 2 \sin \frac{110^{\circ}+20^{\circ}}{2} \sin \frac{110^{\circ}-20^{\circ}}{2} = 2 \sin 65^{\circ} \sin 45^{\circ}$
$= 2 \sin 65^{\circ} \times \frac{1}{\sqrt{2}} = \sqrt{2} \sin 65^{\circ}$
Substituting back:
$\sqrt{2} \sin 65^{\circ} - \sqrt{2} \sin 65^{\circ} + 1 = 1$

(B) Given equations are:
$\tan \theta + \sin \theta = a$ $(1)$
$\tan \theta - \sin \theta = b$ $(2)$
Adding $(1)$ and $(2)$:
$2 \tan \theta = a + b$ $\Rightarrow \tan \theta = \frac{a+b}{2}$ $\Rightarrow \cot \theta = \frac{2}{a+b}$
Subtracting $(2)$ from $(1)$:
$2 \sin \theta = a - b$ $\Rightarrow \sin \theta = \frac{a-b}{2}$ $\Rightarrow \operatorname{cosec} \theta = \frac{2}{a-b}$
Thus,the values are $\frac{2}{a+b}$ and $\frac{2}{a-b}$.

(C) Given: $x = 3 \sin \theta$,$y = 3 \cos \theta \cos \phi$,$z = 3 \cos \theta \sin \phi$
$x^{2} + y^{2} + z^{2} = (3 \sin \theta)^{2} + (3 \cos \theta \cos \phi)^{2} + (3 \cos \theta \sin \phi)^{2}$
$= 9 \sin^{2} \theta + 9 \cos^{2} \theta \cos^{2} \phi + 9 \cos^{2} \theta \sin^{2} \phi$
$= 9 \sin^{2} \theta + 9 \cos^{2} \theta (\cos^{2} \phi + \sin^{2} \phi)$
Since $\cos^{2} \phi + \sin^{2} \phi = 1$:
$= 9 \sin^{2} \theta + 9 \cos^{2} \theta (1)$
$= 9 (\sin^{2} \theta + \cos^{2} \theta)$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$:
$= 9 \times 1 = 9$

(A) Given: $\tan \theta + \cot \theta = 4$
Squaring both sides,we get:
$(\tan \theta + \cot \theta)^{2} = 4^{2}$
$\tan^{2} \theta + \cot^{2} \theta + 2 \tan \theta \cot \theta = 16$
Since $\tan \theta \cot \theta = 1$,we have:
$\tan^{2} \theta + \cot^{2} \theta + 2(1) = 16$
$\tan^{2} \theta + \cot^{2} \theta = 14$
Squaring both sides again:
$(\tan^{2} \theta + \cot^{2} \theta)^{2} = 14^{2}$
$\tan^{4} \theta + \cot^{4} \theta + 2 \tan^{2} \theta \cot^{2} \theta = 196$
$\tan^{4} \theta + \cot^{4} \theta + 2(1)^{2} = 196$
$\tan^{4} \theta + \cot^{4} \theta = 196 - 2 = 194$

(B) Given that $\sin x + \operatorname{cosec} x = 3$.
Squaring both sides,we get:
$(\sin x + \operatorname{cosec} x)^{2} = 3^{2}$
$\sin^{2} x + \operatorname{cosec}^{2} x + 2 \sin x \operatorname{cosec} x = 9$
Since $\sin x \operatorname{cosec} x = 1$,we have:
$\sin^{2} x + \operatorname{cosec}^{2} x + 2(1) = 9$
$\sin^{2} x + \operatorname{cosec}^{2} x = 7$
Now,squaring both sides again:
$(\sin^{2} x + \operatorname{cosec}^{2} x)^{2} = 7^{2}$
$\sin^{4} x + \operatorname{cosec}^{4} x + 2 \sin^{2} x \operatorname{cosec}^{2} x = 49$
$\sin^{4} x + \operatorname{cosec}^{4} x + 2(1)^{2} = 49$
$\sin^{4} x + \operatorname{cosec}^{4} x = 49 - 2 = 47$

(D) Given that $3 \sin \alpha = 5 \sin \beta$,we can write $\frac{\sin \alpha}{\sin \beta} = \frac{5}{3}$.
Applying Componendo and Dividendo,we get:
$\frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{5 + 3}{5 - 3} = \frac{8}{2} = 4$.
Using the sum-to-product formulas:
$\sin \alpha + \sin \beta = 2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$
$\sin \alpha - \sin \beta = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)$
Substituting these into the ratio:
$\frac{2 \sin \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)}{2 \cos \left(\frac{\alpha + \beta}{2}\right) \sin \left(\frac{\alpha - \beta}{2}\right)} = 4$
$\tan \left(\frac{\alpha + \beta}{2}\right) \cot \left(\frac{\alpha - \beta}{2}\right) = 4$
Since $\cot \theta = \frac{1}{\tan \theta}$,we have $\tan \left(\frac{\alpha + \beta}{2}\right) \div \tan \left(\frac{\alpha - \beta}{2}\right) = 4$.

(A) Let $E = \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ}$.
$E = \sqrt{3} \frac{\cos 20^{\circ}}{\sin 20^{\circ}} - 4 \cos 20^{\circ}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - 4 \sin 20^{\circ} \cos 20^{\circ}}{\sin 20^{\circ}}$
Using $2 \sin \theta \cos \theta = \sin 2\theta$,we get $4 \sin 20^{\circ} \cos 20^{\circ} = 2 \sin 40^{\circ}$.
$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 \sin 40^{\circ}}{\sin 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 \sin(60^{\circ} - 20^{\circ})}{\sin 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\sin 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 (\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - \sqrt{3} \cos 20^{\circ} + \sin 20^{\circ}}{\sin 20^{\circ}}$
$E = \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = 1$.

(A) Let $x = \tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ}$.
We know that $\cot 50^{\circ} = \frac{1}{\tan 50^{\circ}}$.
So,$x = \frac{\tan 20^{\circ} \tan 80^{\circ}}{\tan 50^{\circ}}$.
Using the identity $\tan \theta \tan(60^{\circ} - \theta) \tan(60^{\circ} + \theta) = \tan 3\theta$,we have $\tan 20^{\circ} \tan(60^{\circ} - 20^{\circ}) \tan(60^{\circ} + 20^{\circ}) = \tan(3 \times 20^{\circ}) = \tan 60^{\circ} = \sqrt{3}$.
Thus,$\tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ} = \sqrt{3}$.
We need to evaluate $\frac{\tan 20^{\circ} \tan 80^{\circ}}{\tan 50^{\circ}}$.
Using the identity $\tan(60^{\circ} - \theta) \tan \theta \tan(60^{\circ} + \theta) = \tan 3\theta$,we can write $\tan 80^{\circ} = \tan(60^{\circ} + 20^{\circ})$.
Also,$\tan 50^{\circ} = \tan(60^{\circ} - 10^{\circ})$ is not directly helpful,but we know $\tan 50^{\circ} = \tan(60^{\circ} - 10^{\circ})$ and $\tan 70^{\circ} = \tan(60^{\circ} + 10^{\circ})$.
Alternatively,$x = \frac{\sin 20^{\circ} \sin 80^{\circ} \cos 50^{\circ}}{\cos 20^{\circ} \cos 80^{\circ} \sin 50^{\circ}}$.
Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we get $\sin 20^{\circ} \sin 80^{\circ} = \frac{1}{2}(\cos 60^{\circ} - \cos 100^{\circ}) = \frac{1}{2}(\frac{1}{2} + \sin 10^{\circ})$.
After simplification,the value is $\sqrt{3}$.