Trigonometrical ratios of multiple and sub-multiple angles Questions in English

(D) We know that $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.
Substituting these into the expression:
$\sin^2 \theta \cos^2 \theta = \left(\frac{1 - \cos 2\theta}{2}\right) \left(\frac{1 + \cos 2\theta}{2}\right)$
$= \frac{(1 - \cos 2\theta)(1 + \cos 2\theta)}{4}$
$= \frac{1 - \cos^2 2\theta}{4}$
Using the identity $\cos^2 A = \frac{1 + \cos 2A}{2}$,we have $\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}$.
$= \frac{1 - \left(\frac{1 + \cos 4\theta}{2}\right)}{4}$
$= \frac{\frac{2 - 1 - \cos 4\theta}{2}}{4}$
$= \frac{1 - \cos 4\theta}{8}$

(C) We use the identity $\tan \theta - \cot \theta = -2 \cot 2 \theta$.
Starting with the expression: $S = \tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha$.
Using $\tan \alpha = \cot \alpha - 2 \cot 2 \alpha$,we substitute:
$S = (\cot \alpha - 2 \cot 2 \alpha) + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha$
$S = \cot \alpha + 2(\tan 2 \alpha - \cot 2 \alpha) + 4 \tan 4 \alpha + 8 \cot 8 \alpha$
Since $\tan 2 \alpha - \cot 2 \alpha = -2 \cot 4 \alpha$:
$S = \cot \alpha + 2(-2 \cot 4 \alpha) + 4 \tan 4 \alpha + 8 \cot 8 \alpha$
$S = \cot \alpha - 4 \cot 4 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha$
$S = \cot \alpha + 4(\tan 4 \alpha - \cot 4 \alpha) + 8 \cot 8 \alpha$
Since $\tan 4 \alpha - \cot 4 \alpha = -2 \cot 8 \alpha$:
$S = \cot \alpha + 4(-2 \cot 8 \alpha) + 8 \cot 8 \alpha$
$S = \cot \alpha - 8 \cot 8 \alpha + 8 \cot 8 \alpha$
$S = \cot \alpha$.

(B) Given,$\alpha = \frac{180^{\circ}}{7}$,which implies $7 \alpha = 180^{\circ} = \pi$.
We know the identity $3 \sin \alpha - 4 \sin^{3} \alpha = \sin 3 \alpha$.
Substituting the value of $\alpha$:
$\sin 3 \alpha = \sin (7 \alpha - 4 \alpha)$.
Since $7 \alpha = \pi$,we have:
$\sin (\pi - 4 \alpha) = \sin 4 \alpha$.
Thus,$3 \sin \alpha - 4 \sin^{3} \alpha = \sin 4 \alpha$.

(C) We use the hyperbolic identities:
$1$. $\cosh x = 2 \cosh^2 \frac{x}{2} - 1 \implies \cosh \frac{x}{2} = \sqrt{\frac{1+\cosh x}{2}}$
$2$. $\cosh x = 1 + 2 \sinh^2 \frac{x}{2} \implies \sinh \frac{x}{2} = \sqrt{\frac{\cosh x - 1}{2}}$
$3$. $\tanh \frac{x}{2} = \frac{\sinh x}{1+\cosh x} = \frac{\cosh x - 1}{\sinh x}$
$4$. $\sinh x = 2 \sinh \frac{x}{2} \cosh \frac{x}{2}$
Evaluating option $C$: $\frac{\sinh x}{\sqrt{2(\cosh x+1)}} = \frac{2 \sinh \frac{x}{2} \cosh \frac{x}{2}}{\sqrt{2(2 \cosh^2 \frac{x}{2})}} = \frac{2 \sinh \frac{x}{2} \cosh \frac{x}{2}}{2 \cosh \frac{x}{2}} = \sinh \frac{x}{2}$.
Thus,option $C$ is correct.

(B) We are given $\tanh(x) = \frac{1}{3}$.
Using the formula $\tanh(3x) = \frac{3\tanh(x) + \tanh^3(x)}{1 + 3\tanh^2(x)}$:
Substitute $\tanh(x) = \frac{1}{3}$ into the formula:
$\tanh(3x) = \frac{3(\frac{1}{3}) + (\frac{1}{3})^3}{1 + 3(\frac{1}{3})^2}$
$\tanh(3x) = \frac{1 + \frac{1}{27}}{1 + 3(\frac{1}{9})}$
$\tanh(3x) = \frac{\frac{28}{27}}{1 + \frac{1}{3}}$
$\tanh(3x) = \frac{\frac{28}{27}}{\frac{4}{3}}$
$\tanh(3x) = \frac{28}{27} \times \frac{3}{4} = \frac{7}{9}$

(D) Given that $\tan \left(\frac{x}{2}\right) = \frac{m}{n}$.
We use the half-angle formulas: $\sin (x) = \frac{2 \tan (x/2)}{1 + \tan^2 (x/2)}$ and $\cos (x) = \frac{1 - \tan^2 (x/2)}{1 + \tan^2 (x/2)}$.
Substituting these into the expression $m \sin (x) + n \cos (x)$:
$m \left( \frac{2 \tan (x/2)}{1 + \tan^2 (x/2)} \right) + n \left( \frac{1 - \tan^2 (x/2)}{1 + \tan^2 (x/2)} \right)$
$= m \left( \frac{2(m/n)}{1 + (m^2/n^2)} \right) + n \left( \frac{1 - (m^2/n^2)}{1 + (m^2/n^2)} \right)$
$= m \left( \frac{2m/n}{(n^2 + m^2)/n^2} \right) + n \left( \frac{(n^2 - m^2)/n^2}{(n^2 + m^2)/n^2} \right)$
$= \frac{2m^2 n}{n^2 + m^2} + \frac{n(n^2 - m^2)}{n^2 + m^2}$
$= \frac{2m^2 n + n^3 - nm^2}{m^2 + n^2}$
$= \frac{m^2 n + n^3}{m^2 + n^2} = \frac{n(m^2 + n^2)}{m^2 + n^2} = n$.

(A) Given $\cos A = \frac{7}{25}$ and $A$ is in the fourth quadrant,i.e.,$\frac{3 \pi}{2} < A < 2 \pi$.
Since $\frac{3 \pi}{2} < A < 2 \pi$,we have $\frac{3 \pi}{4} < \frac{A}{2} < \pi$ and $\frac{3 \pi}{8} < \frac{A}{4} < \frac{\pi}{2}$.
Using $\cos A = 2 \cos^2 \frac{A}{2} - 1$,we get $\cos^2 \frac{A}{2} = \frac{1 + \cos A}{2} = \frac{1 + 7/25}{2} = \frac{32/25}{2} = \frac{16}{25}$.
Since $\frac{A}{2}$ is in the second quadrant,$\cos \frac{A}{2} = -\frac{4}{5}$.
Using $\cos \frac{A}{2} = 2 \cos^2 \frac{A}{4} - 1$,we get $2 \cos^2 \frac{A}{4} = 1 + \cos \frac{A}{2} = 1 - \frac{4}{5} = \frac{1}{5}$,so $\cos^2 \frac{A}{4} = \frac{1}{10}$.
Since $\frac{A}{4}$ is in the first quadrant,$\cos \frac{A}{4} = \frac{1}{\sqrt{10}}$.
Also,$\cos 2A = 2 \cos^2 A - 1 = 2 \left(\frac{7}{25}\right)^2 - 1 = 2 \left(\frac{49}{625}\right) - 1 = \frac{98 - 625}{625} = -\frac{527}{625}$.
Therefore,$\cos \frac{A}{4} + \cos \frac{A}{2} - \cos 2A = \frac{1}{\sqrt{10}} - \frac{4}{5} - \left(-\frac{527}{625}\right) = \frac{1}{\sqrt{10}} - \frac{500}{625} + \frac{527}{625} = \frac{1}{\sqrt{10}} + \frac{27}{625}$.

(D) Given expression: $32 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) = 16 \left[ 2 \sin \left(\frac{A}{2}\right) \sin \left(\frac{5A}{2}\right) \right]$
Using the formula $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$:
$= 16 \left[ \cos\left(\frac{5A}{2} - \frac{A}{2}\right) - \cos\left(\frac{5A}{2} + \frac{A}{2}\right) \right]$
$= 16 [ \cos(2A) - \cos(3A) ]$
Using $\cos 2A = 2 \cos^2 A - 1$ and $\cos 3A = 4 \cos^3 A - 3 \cos A$:
$= 16 [ (2 \cos^2 A - 1) - (4 \cos^3 A - 3 \cos A) ]$
Substitute $\cos A = \frac{3}{4}$:
$= 16 \left[ 2\left(\frac{3}{4}\right)^2 - 1 - 4\left(\frac{3}{4}\right)^3 + 3\left(\frac{3}{4}\right) \right]$
$= 16 \left[ 2\left(\frac{9}{16}\right) - 1 - 4\left(\frac{27}{64}\right) + \frac{9}{4} \right]$
$= 16 \left[ \frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4} \right]$
$= 16 \left[ \frac{18 - 16 - 27 + 36}{16} \right]$
$= 18 - 16 - 27 + 36 = 11$

(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 $P = \frac{1}{512 \sin(\frac{\pi}{2^{10}})} = \frac{\operatorname{cosec}(\frac{\pi}{2^{10}})}{512}$.

(D) Given,$\sin^4 \theta \cos^2 \theta = \sum_{n=0}^{\infty} a_{2n} \cos 2n \theta$.
We know that $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$.
Substituting these,we get:
$\sin^4 \theta \cos^2 \theta = (\sin^2 \theta)^2 \cos^2 \theta = \left(\frac{1 - \cos 2\theta}{2}\right)^2 \left(\frac{1 + \cos 2\theta}{2}\right)$
$= \frac{1}{8} (1 - 2\cos 2\theta + \cos^2 2\theta)(1 + \cos 2\theta)$
$= \frac{1}{8} (1 + \cos 2\theta - 2\cos 2\theta - 2\cos^2 2\theta + \cos^2 2\theta + \cos^3 2\theta)$
$= \frac{1}{8} (1 - \cos 2\theta - \cos^2 2\theta + \cos^3 2\theta)$
Using $\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}$ and $\cos^3 2\theta = \frac{3\cos 2\theta + \cos 6\theta}{4}$:
$= \frac{1}{8} \left(1 - \cos 2\theta - \frac{1 + \cos 4\theta}{2} + \frac{3\cos 2\theta + \cos 6\theta}{4}\right)$
$= \frac{1}{8} \left(1 - \cos 2\theta - \frac{1}{2} - \frac{1}{2}\cos 4\theta + \frac{3}{4}\cos 2\theta + \frac{1}{4}\cos 6\theta\right)$
$= \frac{1}{8} \left(\frac{1}{2} - \frac{1}{4}\cos 2\theta - \frac{1}{2}\cos 4\theta + \frac{1}{4}\cos 6\theta\right)$
$= \frac{1}{16} - \frac{1}{32}\cos 2\theta - \frac{1}{16}\cos 4\theta + \frac{1}{32}\cos 6\theta$.
Comparing this with $\sum_{n=0}^{\infty} a_{2n} \cos 2n \theta$,we have $a_0 = \frac{1}{16}$,$a_2 = -\frac{1}{32}$,$a_4 = -\frac{1}{16}$,$a_6 = \frac{1}{32}$,and $a_{2n} = 0$ for $n \ge 4$.
The least $n$ for which $a_{2n} = 0$ is $4$.

(C) We use the formula $\prod_{k=1}^{n} \cos \frac{\theta}{2^k} = \frac{\sin \theta}{2^n \sin(\theta/2^n)}$.
Here,$\theta = \frac{\pi}{2}$ and $n=4$.
So,$\cos \frac{\pi}{4} \cos \frac{\pi}{8} \cos \frac{\pi}{16} \cos \frac{\pi}{32} = \frac{\sin(\pi/2)}{2^4 \sin(\pi/32)} = \frac{1}{16 \sin(\pi/32)}$.
This is equal to $\frac{1}{16} \operatorname{cosec} \frac{\pi}{32} = 2^{-4} \operatorname{cosec} \frac{\pi}{32}$.
Comparing with $2^m \operatorname{cosec} \frac{\pi}{n}$,we get $m = -4$ and $n = 32$.
Thus,$m+n = -4 + 32 = 28$.

(B) Given the equation: $\tan \theta + \tan \left(\theta + \frac{\pi}{3}\right) + \tan \left(\theta + \frac{2\pi}{3}\right) = 3$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we expand the terms:
$\tan \theta + \frac{\tan \theta + \sqrt{3}}{1 - \sqrt{3} \tan \theta} + \frac{\tan \theta - \sqrt{3}}{1 + \sqrt{3} \tan \theta} = 3$.
Combining the fractions:
$\tan \theta + \frac{(\tan \theta + \sqrt{3})(1 + \sqrt{3} \tan \theta) + (\tan \theta - \sqrt{3})(1 - \sqrt{3} \tan \theta)}{1 - 3 \tan^2 \theta} = 3$.
Simplifying the numerator:
$\tan \theta + \frac{\tan \theta + \sqrt{3} \tan^2 \theta + \sqrt{3} + 3 \tan \theta + \tan \theta - \sqrt{3} \tan^2 \theta - \sqrt{3} + 3 \tan \theta}{1 - 3 \tan^2 \theta} = 3$.
$\tan \theta + \frac{8 \tan \theta}{1 - 3 \tan^2 \theta} = 3$.
$\frac{\tan \theta - 3 \tan^3 \theta + 8 \tan \theta}{1 - 3 \tan^2 \theta} = 3$.
$\frac{9 \tan \theta - 3 \tan^3 \theta}{1 - 3 \tan^2 \theta} = 3$.
$3 \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) = 3$.
Since $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we get $3 \tan 3\theta = 3$,which implies $\tan 3\theta = 1$.
Thus,option $B$ is correct.

(A) Given $630^{\circ} < \theta < 810^{\circ}$,dividing by $4$ gives $157.5^{\circ} < \frac{\theta}{4} < 202.5^{\circ}$.
Since $\tan \theta = -\frac{7}{24}$ and $\theta$ is in the $4$th quadrant $(630^{\circ} < \theta < 720^{\circ})$ or $1$st quadrant $(720^{\circ} < \theta < 810^{\circ})$,and $\tan \theta < 0$,$\theta$ must be in the $4$th quadrant $(630^{\circ} < \theta < 720^{\circ})$.
Then $\sec^2 \theta = 1 + \tan^2 \theta = 1 + \frac{49}{576} = \frac{625}{576}$,so $\sec \theta = \frac{25}{24}$ (since $\cos \theta > 0$ in the $4$th quadrant).
Thus $\cos \theta = \frac{24}{25}$.
Using the half-angle formula $\cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2} = \frac{1 + 24/25}{2} = \frac{49}{50}$,so $\cos \frac{\theta}{2} = \pm \frac{7}{5 \sqrt{2}}$.
Since $315^{\circ} < \frac{\theta}{2} < 360^{\circ}$,$\cos \frac{\theta}{2} > 0$,so $\cos \frac{\theta}{2} = \frac{7}{5 \sqrt{2}}$.
Then $\cos^2 \frac{\theta}{4} = \frac{1 + \cos \frac{\theta}{2}}{2} = \frac{1 + \frac{7}{5 \sqrt{2}}}{2} = \frac{5 \sqrt{2} + 7}{10 \sqrt{2}}$.
Since $157.5^{\circ} < \frac{\theta}{4} < 202.5^{\circ}$,$\cos \frac{\theta}{4}$ is negative,so $\cos \frac{\theta}{4} = -\sqrt{\frac{7 + 5 \sqrt{2}}{10 \sqrt{2}}}$.

(A) Let $x = \frac{\pi}{8}$. Then $\frac{3\pi}{8} = 3x$.
The expression is $\cos^3 x \cos 3x + \sin^3 x \sin 3x$.
We know that $\cos 3x = 4\cos^3 x - 3\cos x$ and $\sin 3x = 3\sin x - 4\sin^3 x$.
Substituting these:
$\cos^3 x (4\cos^3 x - 3\cos x) + \sin^3 x (3\sin x - 4\sin^3 x) = 4\cos^6 x - 3\cos^4 x + 3\sin^4 x - 4\sin^6 x$.
$= 4(\cos^6 x - \sin^6 x) - 3(\cos^4 x - \sin^4 x)$.
Using $a^2 - b^2 = (a-b)(a+b)$ and $a^3 - b^3 = (a-b)(a^2+ab+b^2)$:
$= 4(\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) - 3(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$.
Since $\cos^2 x + \sin^2 x = 1$,$\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$.
The expression becomes $(\cos^2 x - \sin^2 x) [4(1 - 2\sin^2 x \cos^2 x + \sin^2 x \cos^2 x) - 3]$.
$= \cos 2x [4(1 - \sin^2 x \cos^2 x) - 3] = \cos 2x [1 - 4\sin^2 x \cos^2 x] = \cos 2x [1 - (2\sin x \cos x)^2]$.
$= \cos 2x [1 - \sin^2 2x] = \cos 2x \cdot \cos^2 2x = \cos^3 2x$.
Since $x = \frac{\pi}{8}$,$2x = \frac{\pi}{4}$.
$\cos^3(\frac{\pi}{4}) = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$.

(A) We have the expression $\left(\frac{\sin 3 \theta}{\sin \theta}\right)^2-\left(\frac{\cos 3 \theta}{\cos \theta}\right)^2$.
Using $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$ and $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$,we get:
$\left(\frac{3 \sin \theta - 4 \sin^3 \theta}{\sin \theta}\right)^2 - \left(\frac{4 \cos^3 \theta - 3 \cos \theta}{\cos \theta}\right)^2$
$= (3 - 4 \sin^2 \theta)^2 - (4 \cos^2 \theta - 3)^2$
$= (3 - 4(1 - \cos^2 \theta))^2 - (4 \cos^2 \theta - 3)^2$
$= (4 \cos^2 \theta - 1)^2 - (4 \cos^2 \theta - 3)^2$
Using $x^2 - y^2 = (x - y)(x + y)$:
$= ((4 \cos^2 \theta - 1) - (4 \cos^2 \theta - 3))((4 \cos^2 \theta - 1) + (4 \cos^2 \theta - 3))$
$= (2)(8 \cos^2 \theta - 4) = 16 \cos^2 \theta - 8$
$= 8(2 \cos^2 \theta - 1) = 8 \cos 2 \theta$.
Comparing this with $a \cos b \theta$,we get $a = 8$ and $b = 2$.
Therefore,$a: b = 8: 2 = 4: 1$.

(C) Given $\cos \theta = \frac{-3}{5}$.
Since $\cos \theta < 0$ and $\theta$ is not in the second quadrant,$\theta$ must lie in the third quadrant.
Thus,$\pi < \theta < \frac{3\pi}{2}$.
Dividing by $2$,we get $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$.
This implies $\frac{\theta}{2}$ lies in the second quadrant,where $\tan \frac{\theta}{2}$ is negative.
Using the formula $\tan \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$,we have:
$\tan \frac{\theta}{2} = - \sqrt{\frac{1 - (\frac{-3}{5})}{1 + (\frac{-3}{5})}}$
$= - \sqrt{\frac{1 + \frac{3}{5}}{1 - \frac{3}{5}}}$
$= - \sqrt{\frac{8/5}{2/5}}$
$= - \sqrt{4} = -2$.

(D) Given the equation: $(\sin x)(\cos x) = \frac{1}{4}$
Multiply both sides by $2$: $2 \sin x \cos x = \frac{2}{4}$
Using the identity $\sin 2x = 2 \sin x \cos x$,we get: $\sin 2x = \frac{1}{2}$
Since $x \in \left(0, \frac{\pi}{2}\right)$,then $2x \in (0, \pi)$.
The values of $2x$ for which $\sin 2x = \frac{1}{2}$ are $2x = \frac{\pi}{6}$ and $2x = \frac{5\pi}{6}$.
Therefore,$x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$.
Both values lie in the interval $\left(0, \frac{\pi}{2}\right)$.
Comparing with the given options,$\frac{\pi}{12}$ is the correct value.

(B) Given $\cos^4 \theta = a \cos 4\theta + b \cos 2\theta + c$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have:
$\cos^4 \theta = \left(\frac{1 + \cos 2\theta}{2}\right)^2 = \frac{1}{4} (1 + 2 \cos 2\theta + \cos^2 2\theta)$.
Using $\cos^2 2\theta = \frac{1 + \cos 4\theta}{2}$,we get:
$\cos^4 \theta = \frac{1}{4} + \frac{1}{2} \cos 2\theta + \frac{1}{4} \left(\frac{1 + \cos 4\theta}{2}\right)$.
$= \frac{1}{4} + \frac{1}{2} \cos 2\theta + \frac{1}{8} + \frac{1}{8} \cos 4\theta$.
$= \frac{1}{8} \cos 4\theta + \frac{1}{2} \cos 2\theta + \frac{3}{8}$.
Comparing this with $a \cos 4\theta + b \cos 2\theta + c$,we get $a = \frac{1}{8}$,$b = \frac{1}{2}$,and $c = \frac{3}{8}$.
Thus,$(a, b, c) = \left(\frac{1}{8}, \frac{1}{2}, \frac{3}{8}\right)$.

(A) We have,$\sin (5 \theta) = \sin (3 \theta + 2 \theta)$.
Using the identity $\sin (A + B) = \sin A \cos B + \cos A \sin B$,we get:
$\sin (5 \theta) = \sin 3 \theta \cos 2 \theta + \cos 3 \theta \sin 2 \theta$.
Substituting the multiple angle formulas $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$,$\cos 2 \theta = 2 \cos^2 \theta - 1$,$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$,and $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$\sin (5 \theta) = (3 \sin \theta - 4 \sin^3 \theta)(2 \cos^2 \theta - 1) + (4 \cos^3 \theta - 3 \cos \theta)(2 \sin \theta \cos \theta)$.
Expanding the terms:
$\sin (5 \theta) = (6 \sin \theta \cos^2 \theta - 3 \sin \theta - 8 \sin^3 \theta \cos^2 \theta + 4 \sin^3 \theta) + (8 \sin \theta \cos^4 \theta - 6 \sin \theta \cos^2 \theta)$.
Simplifying by canceling $6 \sin \theta \cos^2 \theta$:
$\sin (5 \theta) = 8 \sin \theta \cos^4 \theta - 8 \sin^3 \theta \cos^2 \theta + 4 \sin^3 \theta - 3 \sin \theta$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$:
$\sin (5 \theta) = 8 \sin \theta \cos^4 \theta - 8 \sin \theta (1 - \cos^2 \theta) \cos^2 \theta + 4 \sin \theta (1 - \cos^2 \theta) - 3 \sin \theta$.
$\sin (5 \theta) = 8 \sin \theta \cos^4 \theta - 8 \sin \theta \cos^2 \theta + 8 \sin \theta \cos^4 \theta + 4 \sin \theta - 4 \sin \theta \cos^2 \theta - 3 \sin \theta$.
$\sin (5 \theta) = 16 \sin \theta \cos^4 \theta - 12 \sin \theta \cos^2 \theta + \sin \theta$.

(D) We know that $\cos 5 \theta = \cos (4 \theta + \theta)$.
Using the formula $\cos (A + B) = \cos A \cos B - \sin A \sin B$,we have:
$\cos 5 \theta = \cos 4 \theta \cos \theta - \sin 4 \theta \sin \theta$
$= (2 \cos ^2 2 \theta - 1) \cos \theta - (2 \sin 2 \theta \cos 2 \theta) \sin \theta$
$= \{2(2 \cos ^2 \theta - 1)^2 - 1\} \cos \theta - 2(2 \sin \theta \cos \theta) \cos 2 \theta \sin \theta$
$= \{2(4 \cos ^4 \theta - 4 \cos ^2 \theta + 1) - 1\} \cos \theta - 4 \sin ^2 \theta \cos \theta (2 \cos ^2 \theta - 1)$
$= (8 \cos ^4 \theta - 8 \cos ^2 \theta + 1) \cos \theta - 4(1 - \cos ^2 \theta) \cos \theta (2 \cos ^2 \theta - 1)$
$= (8 \cos ^5 \theta - 8 \cos ^3 \theta + \cos \theta) - 4 \cos \theta (2 \cos ^2 \theta - 1 - 2 \cos ^4 \theta + \cos ^2 \theta)$
$= 8 \cos ^5 \theta - 8 \cos ^3 \theta + \cos \theta - 4 \cos \theta (3 \cos ^2 \theta - 1 - 2 \cos ^4 \theta)$
$= 8 \cos ^5 \theta - 8 \cos ^3 \theta + \cos \theta - 12 \cos ^3 \theta + 4 \cos \theta + 8 \cos ^5 \theta$
$= 16 \cos ^5 \theta - 20 \cos ^3 \theta + 5 \cos \theta$

(C) We know that $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin(2 \theta)}$.
Here,$\theta = \frac{3 \pi}{16}$,so $2 \theta = \frac{3 \pi}{8}$.
Thus,the expression becomes $\frac{2}{\sin(3 \pi / 8)}$.
Since $\sin(3 \pi / 8) = \cos(\pi / 2 - 3 \pi / 8) = \cos(\pi / 8)$.
Using the half-angle formula,$\cos(\pi / 8) = \sqrt{\frac{1 + \cos(\pi / 4)}{2}} = \sqrt{\frac{1 + 1/\sqrt{2}}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} = \frac{\sqrt{2+\sqrt{2}}}{2}$.
Therefore,$\frac{2}{\sin(3 \pi / 8)} = \frac{2}{\sqrt{2+\sqrt{2}}/2} = \frac{4}{\sqrt{2+\sqrt{2}}}$.
Rationalizing the denominator: $\frac{4}{\sqrt{2+\sqrt{2}}} \times \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}} = \frac{4 \sqrt{2-\sqrt{2}}}{\sqrt{4-2}} = \frac{4 \sqrt{2-\sqrt{2}}}{\sqrt{2}} = 2 \sqrt{2} \sqrt{2-\sqrt{2}} = 2 \sqrt{4-2 \sqrt{2}}$.

(D) We know that $\sin 6 \theta = \sin 3(2 \theta)$.
Using the identity $\sin 3A = 3 \sin A - 4 \sin^3 A$,we get:
$\sin 6 \theta = 3 \sin 2 \theta - 4 \sin^3 2 \theta$.
Substitute $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$\sin 6 \theta = 3(2 \sin \theta \cos \theta) - 4(2 \sin \theta \cos \theta)^3$
$= 6 \sin \theta \cos \theta - 4(8 \sin^3 \theta \cos^3 \theta)$
$= 6 \sin \theta \cos \theta - 32 \sin^3 \theta \cos^3 \theta$
$= 6 \sin \theta \cos \theta - 32 \sin \theta \cos^3 \theta (1 - \cos^2 \theta)$
$= 6 \sin \theta \cos \theta - 32 \sin \theta \cos^3 \theta + 32 \sin \theta \cos^5 \theta$
$= 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3(2 \sin \theta \cos \theta)$
$= 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3 \sin 2 \theta$.
Comparing this with the given equation $\sin 6 \theta = 32 \cos^5 \theta \sin \theta - 32 \cos^3 \theta \sin \theta + 3x$,we find $3x = 3 \sin 2 \theta$,which implies $x = \sin 2 \theta$.

(D) Given $x = \log_e \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$.
By definition,$\sinh x = \frac{e^x - e^{-x}}{2}$.
Here,$e^x = \cot \left( \frac{\pi}{4} + \theta \right) = \frac{1 - \tan \theta}{1 + \tan \theta}$.
Then $e^{-x} = \frac{1}{\cot \left( \frac{\pi}{4} + \theta \right)} = \tan \left( \frac{\pi}{4} + \theta \right) = \frac{1 + \tan \theta}{1 - \tan \theta}$.
Substituting these into the formula for $\sinh x$:
$\sinh x = \frac{1}{2} \left( \frac{1 - \tan \theta}{1 + \tan \theta} - \frac{1 + \tan \theta}{1 - \tan \theta} \right)$
$= \frac{1}{2} \left( \frac{(1 - \tan \theta)^2 - (1 + \tan \theta)^2}{(1 + \tan \theta)(1 - \tan \theta)} \right)$
$= \frac{1}{2} \left( \frac{1 - 2\tan \theta + \tan^2 \theta - (1 + 2\tan \theta + \tan^2 \theta)}{1 - \tan^2 \theta} \right)$
$= \frac{1}{2} \left( \frac{-4\tan \theta}{1 - \tan^2 \theta} \right) = -2 \left( \frac{\tan \theta}{1 - \tan^2 \theta} \right)$.
Since $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$,we have $\sinh x = -\tan 2\theta$.

(C) Let $y = \log \sec x$. Then $\sec x = e^y$,which implies $\cos x = e^{-y}$.
We know that $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
Also,$\cos x = \frac{e^y + e^{-y} - (e^y - e^{-y})}{e^y + e^{-y} + (e^y - e^{-y})} = \frac{2e^{-y}}{2e^y} = e^{-2y}$.
Alternatively,using hyperbolic functions,$\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$.
We have $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} = \frac{\cot^2(x/2) - 1}{\cot^2(x/2) + 1}$.
Using the identity $\coth(y/2) = \frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}}$,we relate this to the expression.
Specifically,$\coth(y/2) = \frac{1 + e^{-y}}{1 - e^{-y}} = \frac{1 + \cos x}{1 - \cos x} = \frac{1 + \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}}{1 - \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}} = \frac{2}{2\tan^2(x/2)} = \cot^2(x/2)$.
Since $\cot^2(x/2) = \operatorname{cosec}^2(x/2) - 1$,we have $\coth(y/2) = \operatorname{cosec}^2(x/2) - 1$.
Therefore,$\frac{y}{2} = \operatorname{coth}^{-1}(\operatorname{cosec}^2(x/2) - 1)$,which gives $y = 2 \operatorname{coth}^{-1}(\operatorname{cosec}^2(x/2) - 1)$.

(D) Given,$90^{\circ} < A < 180^{\circ}$ and $\sin A = \frac{4}{5}$.
Since $90^{\circ} < A < 180^{\circ}$,we have $45^{\circ} < \frac{A}{2} < 90^{\circ}$.
In this interval,$\tan \frac{A}{2}$ must be greater than $1$.
Using the formula $\sin A = \frac{2 \tan \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}$,we have:
$\frac{4}{5} = \frac{2 \tan \frac{A}{2}}{1 + \tan^2 \frac{A}{2}}$
$4(1 + \tan^2 \frac{A}{2}) = 10 \tan \frac{A}{2}$
$4 \tan^2 \frac{A}{2} - 10 \tan \frac{A}{2} + 4 = 0$
$2 \tan^2 \frac{A}{2} - 5 \tan \frac{A}{2} + 2 = 0$
$(2 \tan \frac{A}{2} - 1)(\tan \frac{A}{2} - 2) = 0$
This gives $\tan \frac{A}{2} = \frac{1}{2}$ or $\tan \frac{A}{2} = 2$.
Since $45^{\circ} < \frac{A}{2} < 90^{\circ}$,$\tan \frac{A}{2} > 1$,so $\tan \frac{A}{2} = 2$.

(C) Given $540^{\circ} < \theta < 630^{\circ}$,which implies $\theta$ is in the $3^{rd}$ quadrant.
Since $\tan \theta = \frac{5}{12}$,we have $\sin \theta = -\frac{5}{13}$ and $\cos \theta = -\frac{12}{13}$.
For $\frac{\theta}{2}$,we have $\frac{540^{\circ}}{2} < \frac{\theta}{2} < \frac{630^{\circ}}{2}$,so $270^{\circ} < \frac{\theta}{2} < 315^{\circ}$,which is the $4^{th}$ quadrant.
In the $4^{th}$ quadrant,$\cos \frac{\theta}{2} > 0$ and $\sin \frac{\theta}{2} < 0$.
Using $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$,we get $-\frac{12}{13} = 2 \cos^2 \frac{\theta}{2} - 1$,so $2 \cos^2 \frac{\theta}{2} = \frac{1}{13}$,which gives $\cos \frac{\theta}{2} = \frac{1}{\sqrt{26}}$.
Then $\sin \frac{\theta}{2} = -\sqrt{1 - \cos^2 \frac{\theta}{2}} = -\sqrt{1 - \frac{1}{26}} = -\frac{5}{\sqrt{26}}$.
The denominator is $\sqrt{-(12 \sec \theta + 5 \operatorname{cosec} \theta)} = \sqrt{-(12 \cdot \frac{13}{-12} + 5 \cdot \frac{13}{-5})} = \sqrt{-(-13 - 13)} = \sqrt{26}$.
Substituting these values,the expression becomes $\frac{\frac{1}{\sqrt{26}} - 5(-\frac{5}{\sqrt{26}})}{\sqrt{26}} = \frac{\frac{1+25}{\sqrt{26}}}{\sqrt{26}} = \frac{26}{26} = 1$.

(B) Given $\tan \alpha = \frac{-12}{5}$. Since $\alpha$ is not in the second quadrant,$\alpha$ must be in the fourth quadrant. Thus,$\frac{\alpha}{2} \in (135^{\circ}, 180^{\circ})$.
Using $\tan \alpha = \frac{2 \tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}}$,we get $\frac{-12}{5} = \frac{2 \tan \frac{\alpha}{2}}{1 - \tan^2 \frac{\alpha}{2}}$.
Solving $6 \tan^2 \frac{\alpha}{2} - 5 \tan \frac{\alpha}{2} - 6 = 0$,we find $\tan \frac{\alpha}{2} = \frac{-2}{3}$. Since $\frac{\alpha}{2}$ is in the second quadrant,$\sin \frac{\alpha}{2} = \frac{2}{\sqrt{13}}$.
Given $\cot \beta = \frac{7}{24}$. Since $\beta$ is not in the first quadrant,$\beta$ must be in the third quadrant. Thus,$\cos \beta = \frac{-7}{25}$.
Using $\cos \beta = 2 \cos^2 \frac{\beta}{2} - 1$,we get $\frac{-7}{25} = 2 \cos^2 \frac{\beta}{2} - 1$,so $\cos^2 \frac{\beta}{2} = \frac{9}{25}$. Since $\frac{\beta}{2}$ is in the second quadrant,$\cos \frac{\beta}{2} = \frac{-3}{5}$ and $\sin \frac{\beta}{2} = \frac{4}{5}$,so $\cot \frac{\beta}{2} = \frac{-3}{4}$.
Substituting these values: $\sqrt{13} (\frac{2}{\sqrt{13}}) + (\frac{-3}{5}) + (\frac{-2}{3})(\frac{-3}{4}) = 2 - \frac{3}{5} + \frac{1}{2} = \frac{20 - 6 + 5}{10} = \frac{19}{10}$.

(B) We know that $\cot x = \frac{\cos x}{\sin x}$. Substituting this into the expression: $\frac{\cot ^2 15^{\circ}-1}{\cot ^2 15^{\circ}+1} = \frac{\frac{\cos ^2 15^{\circ}}{\sin ^2 15^{\circ}}-1}{\frac{\cos ^2 15^{\circ}}{\sin ^2 15^{\circ}}+1}$
$= \frac{\cos ^2 15^{\circ}-\sin ^2 15^{\circ}}{\cos ^2 15^{\circ}+\sin ^2 15^{\circ}}$
Using the identities $\cos ^2 x - \sin ^2 x = \cos 2x$ and $\cos ^2 x + \sin ^2 x = 1$:
$= \frac{\cos(2 \times 15^{\circ})}{1} = \cos 30^{\circ}$
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,the final answer is $\frac{\sqrt{3}}{2}$.

(A) Given,$\cot A=\frac{11}{60}$. Since $A$ is not in the first quadrant and $\cot A > 0$,$A$ must be in the third quadrant $(Q_3)$. Thus,$\tan A = \frac{60}{11}$.
Given $\cos B = \frac{7}{25}$. Since $B$ is not in the first quadrant and $\cos B > 0$,$B$ must be in the fourth quadrant $(Q_4)$.
In $Q_4$,$\sin B = -\sqrt{1 - \cos^2 B} = -\sqrt{1 - (\frac{7}{25})^2} = -\frac{24}{25}$.
Thus,$\tan B = \frac{\sin B}{\cos B} = \frac{-24/25}{7/25} = -\frac{24}{7}$.
Using the formula $\tan B = \frac{2 \tan(B/2)}{1 - \tan^2(B/2)}$,we have $-\frac{24}{7} = \frac{2 \tan(B/2)}{1 - \tan^2(B/2)}$.
Let $t = \tan(B/2)$. Then $-\frac{24}{7} = \frac{2t}{1-t^2}$ $\Rightarrow -24 + 24t^2 = 14t$ $\Rightarrow 12t^2 - 7t - 12 = 0$.
Solving the quadratic equation: $12t^2 - 16t + 9t - 12 = 0$ $\Rightarrow 4t(3t - 4) + 3(3t - 4) = 0$ $\Rightarrow (4t + 3)(3t - 4) = 0$.
So,$t = \frac{4}{3}$ or $t = -\frac{3}{4}$.
Since $B \in Q_4$,$\frac{3\pi}{2} < B < 2\pi \Rightarrow \frac{3\pi}{4} < \frac{B}{2} < \pi$. This means $\frac{B}{2}$ is in the second quadrant $(Q_2)$,where $\tan(B/2)$ must be negative. Thus,$\tan(B/2) = -\frac{3}{4}$.
Now,$\tan(A + B/2) = \frac{\tan A + \tan(B/2)}{1 - \tan A \cdot \tan(B/2)} = \frac{\frac{60}{11} - \frac{3}{4}}{1 - (\frac{60}{11})(-\frac{3}{4})} = \frac{\frac{240 - 33}{44}}{1 + \frac{180}{44}} = \frac{207/44}{224/44} = \frac{207}{224}$.
Since $\tan(A + B/2) > 0$,the angle $(A + B/2)$ lies in the first quadrant $(Q_1)$.

(D) We know that $\sin 5 \theta = \sin(3 \theta + 2 \theta) = \sin 3 \theta \cos 2 \theta + \cos 3 \theta \sin 2 \theta$.
Using the identities $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$,$\cos 2 \theta = 1 - 2 \sin^2 \theta$,$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$,and $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$\sin 5 \theta = (3 \sin \theta - 4 \sin^3 \theta)(1 - 2 \sin^2 \theta) + (4 \cos^3 \theta - 3 \cos \theta)(2 \sin \theta \cos \theta)$
$= 3 \sin \theta - 6 \sin^3 \theta - 4 \sin^3 \theta + 8 \sin^5 \theta + 8 \sin \theta \cos^4 \theta - 6 \sin \theta \cos^2 \theta$
$= 8 \sin \theta \cos^4 \theta - 6 \sin \theta \cos^2 \theta + 10 \sin^3 \theta - 10 \sin^5 \theta + 10 \sin^5 \theta$
Alternatively,expanding $\sin 5 \theta$ in terms of $\sin \theta$ and $\cos \theta$:
$\sin 5 \theta = 5 \sin \theta \cos^4 \theta - 10 \sin^3 \theta \cos^2 \theta + \sin^5 \theta$.
Comparing this with $a \cos^4 \theta \sin \theta + b \cos^2 \theta \sin^3 \theta + c \sin^5 \theta$,we get $a = 5$,$b = -10$,and $c = 1$.
Therefore,$abc = 5 \times (-10) \times 1 = -50$.

(C) Given $\cos \theta = \frac{-3}{5}$ and $\pi < \theta < \frac{3\pi}{2}$.
Since $\pi < \theta < \frac{3\pi}{2}$,we have $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$,which lies in the $2^{nd}$ quadrant.
In the $2^{nd}$ quadrant,$\sin \frac{\theta}{2} > 0$,$\cos \frac{\theta}{2} < 0$,and $\tan \frac{\theta}{2} < 0$.
Using the half-angle formulas:
$\cos \theta = 1 - 2 \sin^2 \frac{\theta}{2}$ $\Rightarrow \frac{-3}{5} = 1 - 2 \sin^2 \frac{\theta}{2}$ $\Rightarrow 2 \sin^2 \frac{\theta}{2} = \frac{8}{5}$ $\Rightarrow \sin^2 \frac{\theta}{2} = \frac{4}{5}$.
Since $\sin \frac{\theta}{2} > 0$,we get $\sin \frac{\theta}{2} = \frac{2}{\sqrt{5}}$.
$\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$ $\Rightarrow \frac{-3}{5} = 2 \cos^2 \frac{\theta}{2} - 1$ $\Rightarrow 2 \cos^2 \frac{\theta}{2} = \frac{2}{5}$ $\Rightarrow \cos^2 \frac{\theta}{2} = \frac{1}{5}$.
Since $\cos \frac{\theta}{2} < 0$,we get $\cos \frac{\theta}{2} = -\frac{1}{\sqrt{5}}$.
Then $\tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \frac{2/\sqrt{5}}{-1/\sqrt{5}} = -2$.
Substituting these values into the expression:
$\tan \frac{\theta}{2} + \sin \frac{\theta}{2} + 2 \cos \frac{\theta}{2} = -2 + \frac{2}{\sqrt{5}} + 2 \left( -\frac{1}{\sqrt{5}} \right) = -2 + \frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}} = -2$.

(D) Given $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$.
Squaring both sides: $(\sin \theta - \cos \theta)^2 = (\frac{1}{\sqrt{3}})^2$.
$1 - \sin(2\theta) = \frac{1}{3} \Rightarrow \sin(2\theta) = 1 - \frac{1}{3} = \frac{2}{3}$.
Now,$\cos(4\theta) = 1 - 2\sin^2(2\theta) = 1 - 2(\frac{2}{3})^2 = 1 - 2(\frac{4}{9}) = 1 - \frac{8}{9} = \frac{1}{9}$.
Next,$\sin(6\theta) = 3\sin(2\theta) - 4\sin^3(2\theta) = 3(\frac{2}{3}) - 4(\frac{2}{3})^3 = 2 - 4(\frac{8}{27}) = 2 - \frac{32}{27} = \frac{54 - 32}{27} = \frac{22}{27}$.
Finally,$\sin(2\theta) + \cos(4\theta) + \sin(6\theta) = \frac{2}{3} + \frac{1}{9} + \frac{22}{27} = \frac{18 + 3 + 22}{27} = \frac{43}{27}$.

(C) We use the trigonometric identities: $\cos 2 \theta = 1 - 2 \sin^2 \theta = 2 \cos^2 \theta - 1$ and $\sin 2 \theta = 2 \sin \theta \cos \theta$.
Substituting these into the expression:
$\frac{1 - (1 - 2 \sin^2 \theta) + 2 \sin \theta \cos \theta}{1 + (2 \cos^2 \theta - 1) + 2 \sin \theta \cos \theta}$
$= \frac{2 \sin^2 \theta + 2 \sin \theta \cos \theta}{2 \cos^2 \theta + 2 \sin \theta \cos \theta}$
$= \frac{2 \sin \theta (\sin \theta + \cos \theta)}{2 \cos \theta (\cos \theta + \sin \theta)}$
$= \frac{\sin \theta}{\cos \theta} = \tan \theta$.

(A) We know that $\cot \theta - \tan \theta = \frac{\cos \theta}{\sin \theta} - \frac{\sin \theta}{\cos \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\sin \theta \cos \theta} = \frac{\cos 2 \theta}{\frac{1}{2} \sin 2 \theta} = 2 \cot 2 \theta$.
Substituting this into the expression:
$\cot \theta - \tan \theta - 2 \tan 2 \theta - 4 \tan 4 \theta = 2 \cot 2 \theta - 2 \tan 2 \theta - 4 \tan 4 \theta$.
Now,$2(\cot 2 \theta - \tan 2 \theta) = 2(2 \cot 4 \theta) = 4 \cot 4 \theta$.
Substituting this back:
$4 \cot 4 \theta - 4 \tan 4 \theta = 4(\cot 4 \theta - \tan 4 \theta) = 4(2 \cot 8 \theta) = 8 \cot 8 \theta$.

(B) Given,$3 \sin x + 4 \cos x = 5$.
Using the half-angle formulas $\sin x = \frac{2 \tan(x/2)}{1 + \tan^2(x/2)}$ and $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$,we get:
$3 \left( \frac{2 \tan(x/2)}{1 + \tan^2(x/2)} \right) + 4 \left( \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)} \right) = 5$.
Multiplying both sides by $(1 + \tan^2(x/2))$,we obtain:
$6 \tan(x/2) + 4 - 4 \tan^2(x/2) = 5(1 + \tan^2(x/2))$.
$6 \tan(x/2) + 4 - 4 \tan^2(x/2) = 5 + 5 \tan^2(x/2)$.
Rearranging the terms to isolate the required expression:
$6 \tan(x/2) - 9 \tan^2(x/2) = 5 - 4$.
$6 \tan(x/2) - 9 \tan^2(x/2) = 1$.

(C) We know the identity $\tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3 \tan 3x$.
Given that $\tan x + \tan \left(x + \frac{\pi}{3}\right) + \tan \left(x + \frac{2\pi}{3}\right) = 3$.
Substituting this into the identity,we get $3 \tan 3x = 3$.
Therefore,$\tan 3x = 1$.

(B) Given $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = -\frac{4}{3}$.
$6 \tan A = -4 + 4 \tan^2 A \Rightarrow 2 \tan^2 A - 3 \tan A - 2 = 0$.
Solving for $\tan A$: $(2 \tan A + 1)(\tan A - 2) = 0$,so $\tan A = -\frac{1}{2}$ or $\tan A = 2$.
Since $\tan A < 0$,we have $\tan A = -\frac{1}{2}$.
Now,$\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - 1/4}{1 + 1/4} = \frac{3/4}{5/4} = \frac{3}{5}$.
Using the identity $\cos 6A = 4 \cos^3 2A - 3 \cos 2A$:
$\cos 6A = 4 \left(\frac{3}{5}\right)^3 - 3 \left(\frac{3}{5}\right) = 4 \left(\frac{27}{125}\right) - \frac{9}{5} = \frac{108}{125} - \frac{225}{125} = -\frac{117}{125}$.

(C) We know the trigonometric identity: $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$.
By substituting $\theta = 15^{\circ}$,we get:
$\frac{1-\tan^2 15^{\circ}}{1+\tan^2 15^{\circ}} = \cos(2 \times 15^{\circ})$
$= \cos 30^{\circ}$
$= \frac{\sqrt{3}}{2}$.

(B) Given,$\theta \in \left(\pi, \frac{3 \pi}{2}\right)$.
Since $\pi < \theta < \frac{3 \pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3 \pi}{4}$.
In this interval,$\tan \left(\frac{\theta}{2}\right)$ is negative.
Using the formula $\cos \theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)}$,we have $\frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{-3}{5}$.
Cross-multiplying gives $5 - 5 \tan^2(\theta/2) = -3 - 3 \tan^2(\theta/2)$.
Rearranging terms,$8 = 2 \tan^2(\theta/2)$,which implies $\tan^2(\theta/2) = 4$.
Thus,$\tan(\theta/2) = \pm 2$.
Since $\frac{\theta}{2}$ lies in the second quadrant,$\tan(\theta/2)$ must be negative.
Therefore,$\tan(\theta/2) = -2$.

(A) Given $y = \frac{2 \sin \theta}{1+\cos \theta+\sin \theta}$.
Using half-angle formulas $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$,$\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$,and $1 = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}$:
$y = \frac{2(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}{2 \cos^2 \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = \frac{4 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos \frac{\theta}{2}(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})} = \frac{2 \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}$.
Now,consider the expression $E = \frac{1-\cos \theta+\sin \theta}{1+\sin \theta}$.
Using $1-\cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $1+\sin \theta = (\sin \frac{\theta}{2} + \cos \frac{\theta}{2})^2$:
$E = \frac{2 \sin^2 \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{(\sin \frac{\theta}{2} + \cos \frac{\theta}{2})^2} = \frac{2 \sin \frac{\theta}{2}(\sin \frac{\theta}{2} + \cos \frac{\theta}{2})}{(\sin \frac{\theta}{2} + \cos \frac{\theta}{2})^2} = \frac{2 \sin \frac{\theta}{2}}{\sin \frac{\theta}{2} + \cos \frac{\theta}{2}} = y$.

(C) We have $\operatorname{cosec} 2A + \cot 2A = \frac{1}{\sin 2A} + \frac{\cos 2A}{\sin 2A} = \frac{1 + \cos 2A}{\sin 2A}$.
Using the identities $1 + \cos 2A = 2 \cos^2 A$ and $\sin 2A = 2 \sin A \cos A$,we get:
$\frac{2 \cos^2 A}{2 \sin A \cos A} = \frac{\cos A}{\sin A} = \cot A$.
Now,we express $\cot A$ in terms of $\tan A$:
$\cot A = \frac{1}{\tan A} = \frac{\tan^2 A + 1 - \tan^2 A}{\tan A} = \tan A + \frac{1 - \tan^2 A}{\tan A}$.
Since $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$,we have $\cot 2A = \frac{1 - \tan^2 A}{2 \tan A}$,which implies $\frac{1 - \tan^2 A}{\tan A} = 2 \cot 2A$.
Therefore,$\operatorname{cosec} 2A + \cot 2A = \tan A + 2 \cot 2A$.

(A) We know that $\sin 5 \theta = 5 \sin \theta - 20 \sin ^3 \theta + 16 \sin ^5 \theta$.
Dividing by $\sin \theta$ (assuming $\sin \theta \neq 0$):
$\frac{\sin 5 \theta}{\sin \theta} = 5 - 20 \sin ^2 \theta + 16 \sin ^4 \theta$.
Substitute $\sin ^2 \theta = 1 - \cos ^2 \theta$:
$= 5 - 20(1 - \cos ^2 \theta) + 16(1 - \cos ^2 \theta)^2$
$= 5 - 20 + 20 \cos ^2 \theta + 16(1 - 2 \cos ^2 \theta + \cos ^4 \theta)$
$= -15 + 20 \cos ^2 \theta + 16 - 32 \cos ^2 \theta + 16 \cos ^4 \theta$
$= 16 \cos ^4 \theta - 12 \cos ^2 \theta + 1$.

(A) Let $P = \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)$.
Multiply and divide by $2 \sin \left(\frac{\pi}{7}\right)$:
$P = \frac{2 \sin \left(\frac{\pi}{7}\right) \cos \left(\frac{\pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)}{2 \sin \left(\frac{\pi}{7}\right)}$
Using $2 \sin \theta \cos \theta = \sin 2 \theta$:
$P = \frac{\sin \left(\frac{2 \pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)}{2 \sin \left(\frac{\pi}{7}\right)}$
Multiply numerator and denominator by $2$:
$P = \frac{2 \sin \left(\frac{2 \pi}{7}\right) \cos \left(\frac{2 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)}{4 \sin \left(\frac{\pi}{7}\right)} = \frac{\sin \left(\frac{4 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)}{4 \sin \left(\frac{\pi}{7}\right)}$
Multiply numerator and denominator by $2$ again:
$P = \frac{2 \sin \left(\frac{4 \pi}{7}\right) \cos \left(\frac{4 \pi}{7}\right)}{8 \sin \left(\frac{\pi}{7}\right)} = \frac{\sin \left(\frac{8 \pi}{7}\right)}{8 \sin \left(\frac{\pi}{7}\right)}$
Since $\sin \left(\frac{8 \pi}{7}\right) = \sin \left(\pi + \frac{\pi}{7}\right) = -\sin \left(\frac{\pi}{7}\right)$:
$P = \frac{-\sin \left(\frac{\pi}{7}\right)}{8 \sin \left(\frac{\pi}{7}\right)} = -\frac{1}{8}$.

(A) To evaluate the product $P = \cos A \cos 2 A \cos 2^2 A \ldots \cos 2^{n-1} A$,multiply and divide by $2 \sin A$:
$P = \frac{1}{2 \sin A} (2 \sin A \cos A) \cos 2 A \cos 4 A \ldots \cos 2^{n-1} A$
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we get:
$P = \frac{1}{2 \sin A} (\sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A$
Multiply and divide by $2$:
$P = \frac{1}{2^2 \sin A} (2 \sin 2 A \cos 2 A) \cos 4 A \ldots \cos 2^{n-1} A = \frac{\sin 4 A}{2^2 \sin A} \cos 4 A \ldots \cos 2^{n-1} A$
Continuing this process $n$ times,we obtain:
$P = \frac{\sin 2^n A}{2^n \sin A}$