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

(D) Given $2 \cos \theta = x + \frac{1}{x}$,which implies $\cos \theta = \frac{1}{2} \left(x + \frac{1}{x}\right)$.
Using the triple angle identity $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$,we have:
$2 \cos 3 \theta = 2 [4 \cos^3 \theta - 3 \cos \theta] = 8 \cos^3 \theta - 6 \cos \theta$.
Substitute $\cos \theta = \frac{1}{2} \left(x + \frac{1}{x}\right)$:
$2 \cos 3 \theta = 8 \left[ \frac{1}{2} \left(x + \frac{1}{x}\right) \right]^3 - 6 \left[ \frac{1}{2} \left(x + \frac{1}{x}\right) \right]$
$= 8 \left[ \frac{1}{8} \left(x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) \right) \right] - 3 \left(x + \frac{1}{x}\right)$
$= \left(x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) \right) - 3 \left(x + \frac{1}{x}\right)$
$= x^3 + \frac{1}{x^3}$.

(B) Given $a \sin \theta = b \cos \theta$,we have $\tan \theta = \frac{b}{a}$.
We need to evaluate $a \cos 2 \theta + b \sin 2 \theta$.
Using the double angle formulas in terms of $\tan \theta$:
$a \cos 2 \theta + b \sin 2 \theta = a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)$
Substitute $\tan \theta = \frac{b}{a}$:
$= a \left( \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} \right) + b \left( \frac{2 \left( \frac{b}{a} \right)}{1 + \frac{b^2}{a^2}} \right)$
$= a \left( \frac{a^2 - b^2}{a^2 + b^2} \right) + b \left( \frac{2b}{a} \cdot \frac{a^2}{a^2 + b^2} \right)$
$= \frac{a(a^2 - b^2)}{a^2 + b^2} + \frac{2ab^2}{a^2 + b^2}$
$= \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2} = \frac{a^3 + ab^2}{a^2 + b^2}$
$= \frac{a(a^2 + b^2)}{a^2 + b^2} = a$.

(B) Given $3 \sin \theta = 2 \sin 3 \theta$.
Using the identity $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$,we have:
$3 \sin \theta = 2(3 \sin \theta - 4 \sin^3 \theta)$
$3 \sin \theta = 6 \sin \theta - 8 \sin^3 \theta$
$8 \sin^3 \theta - 3 \sin \theta = 0$
$\sin \theta (8 \sin^2 \theta - 3) = 0$
This implies $\sin \theta = 0$ or $\sin^2 \theta = \frac{3}{8}$.
Since $0 < \theta < \pi$,$\sin \theta$ must be positive and non-zero.
Therefore,$\sin \theta = \sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{2 \sqrt{2}}$.

(A) Given: $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3}$.
First,calculate $\tan 2B$ using the formula $\tan 2B = \frac{2 \tan B}{1 - \tan^2 B}$.
$\tan 2B = \frac{2(\frac{1}{3})}{1 - (\frac{1}{3})^2} = \frac{\frac{2}{3}}{1 - \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$.
Now,use the formula $\tan (A + 2B) = \frac{\tan A + \tan 2B}{1 - \tan A \cdot \tan 2B}$.
$\tan (A + 2B) = \frac{\frac{1}{2} + \frac{3}{4}}{1 - (\frac{1}{2} \times \frac{3}{4})} = \frac{\frac{2+3}{4}}{1 - \frac{3}{8}} = \frac{\frac{5}{4}}{\frac{5}{8}} = \frac{5}{4} \times \frac{8}{5} = 2$.

(A) Given the equation $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$: $2 \sin x \cos x = 2 \times \frac{1}{4} = \frac{1}{2}$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get $\sin 2x = \frac{1}{2}$.
Since $\sin 2x = \sin \frac{\pi}{6}$,the general solution for $2x$ is $2x = n\pi + (-1)^n \frac{\pi}{6}$.
For $n=0$,$2x = \frac{\pi}{6} \implies x = \frac{\pi}{12}$.
For $n=1$,$2x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies x = \frac{5\pi}{12}$.
Both values $x = \frac{\pi}{12}$ and $x = \frac{5\pi}{12}$ lie in the interval $\left(0, \frac{\pi}{2}\right)$.

(B) Given $\sec x = \frac{25}{24}$,we have $\cos x = \frac{24}{25}$.
Since $x$ lies in the first quadrant,$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{24}{25})^2} = \sqrt{\frac{625 - 576}{625}} = \sqrt{\frac{49}{625}} = \frac{7}{25}$.
Now,consider $(\sin \frac{x}{2} + \cos \frac{x}{2})^2 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = 1 + \sin x$.
Substituting the value of $\sin x$,we get $(\sin \frac{x}{2} + \cos \frac{x}{2})^2 = 1 + \frac{7}{25} = \frac{32}{25}$.
Since $x$ is in the first quadrant,$0 < x < \frac{\pi}{2}$,so $0 < \frac{x}{2} < \frac{\pi}{4}$. In this interval,both $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ are positive.
Therefore,$\sin \frac{x}{2} + \cos \frac{x}{2} = \sqrt{\frac{32}{25}} = \frac{\sqrt{16 \times 2}}{5} = \frac{4 \sqrt{2}}{5} = \frac{4 \sqrt{2} \times \sqrt{2}}{5 \sqrt{2}} = \frac{8}{5 \sqrt{2}}$.

(C) We know that $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$.
Setting $\theta = \frac{\pi}{8}$,we have $\tan \frac{\pi}{4} = \frac{2\tan \frac{\pi}{8}}{1-\tan^2 \frac{\pi}{8}}$.
Since $\tan \frac{\pi}{4} = 1$,let $y = \tan \frac{\pi}{8}$.
Then $1 = \frac{2y}{1-y^2}$,which implies $1 - y^2 = 2y$,or $y^2 + 2y - 1 = 0$.
Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $y = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$.
Since $\frac{\pi}{8}$ is in the $1^{\text{st}}$ quadrant,$\tan \frac{\pi}{8} > 0$.
Therefore,$\tan \frac{\pi}{8} = \sqrt{2} - 1$.

(C) Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we have:
$\frac{\tan 160^{\circ} - \tan 110^{\circ}}{1 + \tan 160^{\circ} \tan 110^{\circ}} = \tan(160^{\circ} - 110^{\circ}) = \tan 50^{\circ}$
Since $\tan 50^{\circ} = \tan(90^{\circ} - 40^{\circ}) = \cot 40^{\circ} = \frac{1}{\tan 40^{\circ}}$
Using the double angle formula $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$,we substitute $\theta = 20^{\circ}$:
$\tan 40^{\circ} = \frac{2\tan 20^{\circ}}{1 - \tan^2 20^{\circ}} = \frac{2p}{1 - p^2}$
Therefore,$\frac{1}{\tan 40^{\circ}} = \frac{1 - p^2}{2p}$.

(D) Given $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3 \pi}{2}$.
Since $x$ is in the third quadrant,$\sec^2 x = 1 + \tan^2 x = 1 + \frac{9}{16} = \frac{25}{16}$.
Thus,$\sec x = -\frac{5}{4}$ (as $\sec x < 0$ in the third quadrant),which implies $\cos x = -\frac{4}{5}$.
Given $\pi < x < \frac{3 \pi}{2}$,we have $\frac{\pi}{2} < \frac{x}{2} < \frac{3 \pi}{4}$.
In the interval $(\frac{\pi}{2}, \frac{3 \pi}{4})$,$\cos \frac{x}{2}$ is negative.
Using the half-angle formula,$\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}}$.
Substituting $\cos x = -\frac{4}{5}$,we get $\cos \frac{x}{2} = -\sqrt{\frac{1 - 4/5}{2}} = -\sqrt{\frac{1/5}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

(B) We know that $3A = 2A + A$.
Taking tangent on both sides:
$\tan 3A = \tan(2A + A) = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}$.
Cross-multiplying gives:
$\tan 3A(1 - \tan 2A \tan A) = \tan 2A + \tan A$.
$\tan 3A - \tan 3A \tan 2A \tan A = \tan 2A + \tan A$.
Rearranging the terms to isolate the product:
$\tan 3A \tan 2A \tan A = \tan 3A - \tan 2A - \tan A$.

(D) Given the equation: $3 \sin 2 \theta = 2 \sin 3 \theta$.
Using the identities $\sin 2 \theta = 2 \sin \theta \cos \theta$ and $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$:
$3(2 \sin \theta \cos \theta) = 2(3 \sin \theta - 4 \sin^3 \theta)$.
$6 \sin \theta \cos \theta = 6 \sin \theta - 8 \sin^3 \theta$.
Since $0 < \theta < \pi$,$\sin \theta \neq 0$,so we can divide by $2 \sin \theta$:
$3 \cos \theta = 3 - 4 \sin^2 \theta$.
Substitute $\sin^2 \theta = 1 - \cos^2 \theta$:
$3 \cos \theta = 3 - 4(1 - \cos^2 \theta) = 3 - 4 + 4 \cos^2 \theta$.
$4 \cos^2 \theta - 3 \cos \theta - 1 = 0$.
Factoring the quadratic: $(4 \cos \theta + 1)(\cos \theta - 1) = 0$.
This gives $\cos \theta = 1$ (which implies $\theta = 0$,not in range) or $\cos \theta = -\frac{1}{4}$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$:
$\sin^2 \theta = 1 - (-\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}$.
Therefore,$\sin \theta = \frac{\sqrt{15}}{4}$.

(B) We know that $\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$.
Let $\theta = 10^{\circ}$,then $\tan 30^{\circ} = \frac{1}{\sqrt{3}} = \frac{3\tan 10^{\circ} - \tan^3 10^{\circ}}{1 - 3\tan^2 10^{\circ}}$.
Squaring both sides,we get $\frac{1}{3} = \frac{(3\tan 10^{\circ} - \tan^3 10^{\circ})^2}{(1 - 3\tan^2 10^{\circ})^2}$.
$(1 - 3\tan^2 10^{\circ})^2 = 3(9\tan^2 10^{\circ} - 6\tan^4 10^{\circ} + \tan^6 10^{\circ})$.
$1 - 6\tan^2 10^{\circ} + 9\tan^4 10^{\circ} = 27\tan^2 10^{\circ} - 18\tan^4 10^{\circ} + 3\tan^6 10^{\circ}$.
Rearranging the terms: $3\tan^6 10^{\circ} - 27\tan^4 10^{\circ} + 33\tan^2 10^{\circ} = 1$.

(B) We know that $\cos(\pi - \theta) = -\cos \theta$,so $\cos^4(\pi - \theta) = \cos^4 \theta$.
Thus,$\cos^4 \frac{7\pi}{8} = \cos^4(\pi - \frac{\pi}{8}) = \cos^4 \frac{\pi}{8}$ and $\cos^4 \frac{5\pi}{8} = \cos^4(\pi - \frac{3\pi}{8}) = \cos^4 \frac{3\pi}{8}$.
The expression becomes $2(\cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8})$.
Using $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we have $\cos^4 \theta = (\frac{1 + \cos 2\theta}{2})^2 = \frac{1 + 2\cos 2\theta + \cos^2 2\theta}{4} = \frac{1 + 2\cos 2\theta + \frac{1 + \cos 4\theta}{2}}{4} = \frac{3 + 4\cos 2\theta + \cos 4\theta}{8}$.
For $\theta = \frac{\pi}{8}$,$\cos^4 \frac{\pi}{8} = \frac{3 + 4\cos(\pi/4) + \cos(\pi/2)}{8} = \frac{3 + 4(1/\sqrt{2}) + 0}{8} = \frac{3 + 2\sqrt{2}}{8}$.
For $\theta = \frac{3\pi}{8}$,$\cos^4 \frac{3\pi}{8} = \frac{3 + 4\cos(3\pi/4) + \cos(3\pi/2)}{8} = \frac{3 + 4(-1/\sqrt{2}) + 0}{8} = \frac{3 - 2\sqrt{2}}{8}$.
Summing these,$2(\frac{3 + 2\sqrt{2}}{8} + \frac{3 - 2\sqrt{2}}{8}) = 2(\frac{6}{8}) = 2(\frac{3}{4}) = \frac{3}{2}$.

(D) We know the formula for $\cos 2 \theta$ in terms of $\tan \theta$ is:
$\cos 2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$
Given $\tan \theta = \frac{1}{3}$,substitute this value into the formula:
$\cos 2 \theta = \frac{1 - (\frac{1}{3})^2}{1 + (\frac{1}{3})^2} = \frac{1 - \frac{1}{9}}{1 + \frac{1}{9}}$
$\cos 2 \theta = \frac{\frac{8}{9}}{\frac{10}{9}} = \frac{8}{10} = \frac{4}{5}$

(A) We know that $1 + \cos 2A = 2 \cos^2 A$.
Using this identity,we simplify the expression step by step:
$\sqrt{2+\sqrt{2+2 \cos 4 \theta}} = \sqrt{2+\sqrt{2(1+\cos 4 \theta)}}$
$= \sqrt{2+\sqrt{2(2 \cos^2 2 \theta)}}$
$= \sqrt{2+\sqrt{4 \cos^2 2 \theta}}$
$= \sqrt{2+2 \cos 2 \theta}$
$= \sqrt{2(1+\cos 2 \theta)}$
$= \sqrt{2(2 \cos^2 \theta)}$
$= \sqrt{4 \cos^2 \theta} = 2 \cos \theta$

(A) We know that $\operatorname{cosec} 2 \theta = \frac{1}{\sin 2 \theta}$ and $\cot 2 \theta = \frac{\cos 2 \theta}{\sin 2 \theta}$.
Substituting these in the expression:
$\operatorname{cosec} 2 \theta - \cot 2 \theta = \frac{1}{\sin 2 \theta} - \frac{\cos 2 \theta}{\sin 2 \theta} = \frac{1 - \cos 2 \theta}{\sin 2 \theta}$.
Using the trigonometric identities $1 - \cos 2 \theta = 2 \sin^2 \theta$ and $\sin 2 \theta = 2 \sin \theta \cos \theta$:
$= \frac{2 \sin^2 \theta}{2 \sin \theta \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$.
Thus,the correct option is $A$.

(D) We use the identity $\sin^{2}(\theta) = \frac{1 - \cos(2\theta)}{2}$.
Substituting $\theta = \frac{\pi}{8}$:
$\sin^{2}\left(\frac{\pi}{8}\right) = \frac{1 - \cos\left(2 \times \frac{\pi}{8}\right)}{2}$
$= \frac{1 - \cos\left(\frac{\pi}{4}\right)}{2}$
Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,we have:
$= \frac{1 - \frac{1}{\sqrt{2}}}{2} = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2} = \frac{\sqrt{2}-1}{2\sqrt{2}}$.

(A) We know the formula $\tan \left(\frac{A}{2}\right) = \frac{1-\cos A}{\sin A}$.
Putting $A = \frac{\pi}{4}$,we get $\frac{A}{2} = \frac{\pi}{8}$.
Therefore,$\tan \left(\frac{\pi}{8}\right) = \frac{1-\cos(\pi/4)}{\sin(\pi/4)}$.
Substituting the values $\cos(\pi/4) = \frac{1}{\sqrt{2}}$ and $\sin(\pi/4) = \frac{1}{\sqrt{2}}$,we get:
$\tan \left(\frac{\pi}{8}\right) = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2}-1$.

(A) We know that $\cos 2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$.
Substituting these into the expression $b \cos 2 \theta + a \sin 2 \theta$:
$= b \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + a \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)$
Given $\tan \theta = \frac{a}{b}$,we substitute this value:
$= b \left( \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} \right) + a \left( \frac{2 \frac{a}{b}}{1 + \frac{a^2}{b^2}} \right)$
$= b \left( \frac{\frac{b^2 - a^2}{b^2}}{\frac{b^2 + a^2}{b^2}} \right) + \frac{2a^2}{b \left( \frac{b^2 + a^2}{b^2} \right)}$
$= b \left( \frac{b^2 - a^2}{b^2 + a^2} \right) + \frac{2a^2 b}{b^2 + a^2}$
$= \frac{b^3 - a^2 b + 2a^2 b}{b^2 + a^2}$
$= \frac{b^3 + a^2 b}{b^2 + a^2} = \frac{b(b^2 + a^2)}{b^2 + a^2} = b$.

(B) We use the identity $\cot \theta - \tan \theta = 2 \cot 2\theta$,which implies $\cot \theta = \tan \theta + 2 \cot 2\theta$.
Rearranging,we get $2 \cot 2\theta = \cot \theta - \tan \theta$.
Now,consider the expression $E = \tan A + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A$.
Using the identity $\tan \theta = \cot \theta - 2 \cot 2\theta$,we have:
$\tan A = \cot A - 2 \cot 2A$
Substituting this into the expression:
$E = (\cot A - 2 \cot 2A) + 2 \tan 2A + 4 \tan 4A + 8 \cot 8A$
This approach is complex,so let us simplify step-by-step using $\tan \theta = \cot \theta - 2 \cot 2\theta$ repeatedly:
$8 \cot 8A = 4 \cot 4A - 4 \tan 4A$
$E = \tan A + 2 \tan 2A + 4 \tan 4A + (4 \cot 4A - 4 \tan 4A) = \tan A + 2 \tan 2A + 4 \cot 4A$
Now,$4 \cot 4A = 2 \cot 2A - 2 \tan 2A$
$E = \tan A + 2 \tan 2A + (2 \cot 2A - 2 \tan 2A) = \tan A + 2 \cot 2A$
Finally,$2 \cot 2A = \cot A - \tan A$
$E = \tan A + (\cot A - \tan A) = \cot A$.

(B) We use the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$.
$\cos x \cos 7 x - \cos 5 x \cos 13 x = \frac{1}{2} [2 \cos 7 x \cos x - 2 \cos 13 x \cos 5 x]$
$= \frac{1}{2} [(\cos 8 x + \cos 6 x) - (\cos 18 x + \cos 8 x)]$
$= \frac{1}{2} [\cos 6 x - \cos 18 x]$
Using the formula $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$:
$= \frac{1}{2} [-2 \sin \frac{6 x + 18 x}{2} \sin \frac{6 x - 18 x}{2}]$
$= - \sin 12 x \sin(-6 x)$
$= \sin 12 x \sin 6 x$
Since $\sin 12 x = 2 \sin 6 x \cos 6 x$,we have:
$= (2 \sin 6 x \cos 6 x) \sin 6 x = 2 \sin ^{2} 6 x \cos 6 x$.

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

(B) Given the function $f(x) = 3 \sin x - 4 \sin^3 x$.
We know the trigonometric identity $\sin(3x) = 3 \sin x - 4 \sin^3 x$.
Therefore,$f(x) = \sin(3x)$.
Now,substitute $x = \frac{\pi}{3}$ into the function:
$f\left(\frac{\pi}{3}\right) = \sin\left(3 \times \frac{\pi}{3}\right) = \sin(\pi)$.
Since $\sin(\pi) = 0$,the value is $0$.
Thus,the correct option is $B$.

(C) Given,$\sin x - \sin y = \frac{1}{2} \dots (i)$ and $\cos x - \cos y = 1 \dots (ii)$.
Using sum-to-product formulas:
$2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) = \frac{1}{2} \dots (iii)$
$-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) = 1 \dots (iv)$
Dividing $(iv)$ by $(iii)$:
$\frac{-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)} = \frac{1}{1/2}$
$-\tan \left(\frac{x+y}{2}\right) = 2 \implies \tan \left(\frac{x+y}{2}\right) = -2$.
Using the formula $\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$ where $\theta = \frac{x+y}{2}$:
$\tan(x+y) = \frac{2 \tan \left(\frac{x+y}{2}\right)}{1 - \tan^2 \left(\frac{x+y}{2}\right)} = \frac{2(-2)}{1 - (-2)^2} = \frac{-4}{1 - 4} = \frac{-4}{-3} = \frac{4}{3}$.

(D) We know that $\sin 30^{\circ} = \frac{1}{2}$.
Let $E = \sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}$.
$E = \frac{1}{2} \cdot (\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ})$.
Using the identity $\sin \theta \cdot \sin(60^{\circ}-\theta) \cdot \sin(60^{\circ}+\theta) = \frac{1}{4} \sin 3\theta$,where $\theta = 10^{\circ}$:
$\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ} = \sin 10^{\circ} \cdot \sin(60^{\circ}-10^{\circ}) \cdot \sin(60^{\circ}+10^{\circ}) = \frac{1}{4} \sin(3 \times 10^{\circ}) = \frac{1}{4} \sin 30^{\circ} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$.
Therefore,$E = \frac{1}{2} \cdot \frac{1}{8} = \frac{1}{16}$.

(B) Let $y = \sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}$.
We know that $1 + \cos 2A = 2 \cos^2 A$.
Using this identity,we simplify the expression step by step:
$y = \sqrt{2+\sqrt{2+\sqrt{2(1 + \cos 8 \theta)}}}$
$y = \sqrt{2+\sqrt{2+\sqrt{2(2 \cos^2 4 \theta)}}}$
$y = \sqrt{2+\sqrt{2+\sqrt{4 \cos^2 4 \theta}}}$
$y = \sqrt{2+\sqrt{2+2 \cos 4 \theta}}$
$y = \sqrt{2+\sqrt{2(1 + \cos 4 \theta)}}$
$y = \sqrt{2+\sqrt{2(2 \cos^2 2 \theta)}}$
$y = \sqrt{2+\sqrt{4 \cos^2 2 \theta}}$
$y = \sqrt{2+2 \cos 2 \theta}$
$y = \sqrt{2(1 + \cos 2 \theta)}$
$y = \sqrt{2(2 \cos^2 \theta)}$
$y = \sqrt{4 \cos^2 \theta} = 2 \cos \theta$.

(C) We know the formula $\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$.
Setting $\theta = \frac{\pi}{4}$,we get $\tan \frac{\pi}{8} = \frac{\sin(\pi/4)}{1 + \cos(\pi/4)}$.
Substituting the values $\sin(\pi/4) = \frac{1}{\sqrt{2}}$ and $\cos(\pi/4) = \frac{1}{\sqrt{2}}$,we have:
$\tan \frac{\pi}{8} = \frac{1/\sqrt{2}}{1 + 1/\sqrt{2}} = \frac{1/\sqrt{2}}{(\sqrt{2}+1)/\sqrt{2}} = \frac{1}{\sqrt{2}+1}$.

(C) Given,$\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$.
Using the identities $1+\cos A = 2\cos^{2}\frac{A}{2}$ and $1-\cos A = 2\sin^{2}\frac{A}{2}$,we get:
$\sqrt{\frac{2\cos^{2}\frac{A}{2}}{2\sin^{2}\frac{A}{2}}} = \frac{x}{y}$
$\cot\frac{A}{2} = \frac{x}{y} \implies \tan\frac{A}{2} = \frac{y}{x}$.
Now,using the formula $\tan A = \frac{2\tan\frac{A}{2}}{1-\tan^{2}\frac{A}{2}}$:
$\tan A = \frac{2(\frac{y}{x})}{1-(\frac{y}{x})^{2}} = \frac{\frac{2y}{x}}{\frac{x^{2}-y^{2}}{x^{2}}}$
$\tan A = \frac{2y}{x} \times \frac{x^{2}}{x^{2}-y^{2}} = \frac{2xy}{x^{2}-y^{2}}$.

(C) Given,$\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$.
Since $\pi < x < \frac{3\pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$,which means $\frac{x}{2}$ lies in the second quadrant.
In the second quadrant,$\cos \theta$ is negative.
We know that $\sec^2 x = 1 + \tan^2 x = 1 + (\frac{3}{4})^2 = 1 + \frac{9}{16} = \frac{25}{16}$.
Therefore,$\sec x = \pm \frac{5}{4}$.
Since $x$ is in the third quadrant,$\cos x$ must be negative,so $\cos x = -\frac{4}{5}$.
Using the half-angle formula,$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$.
Substituting the value of $\cos x$,we get $\cos^2 \frac{x}{2} = \frac{1 - \frac{4}{5}}{2} = \frac{1/5}{2} = \frac{1}{10}$.
Since $\frac{x}{2}$ is in the second quadrant,$\cos \frac{x}{2}$ must be negative.
Thus,$\cos \frac{x}{2} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

(D) Given: $\sin \theta = \frac{3}{5}$.
Since $\theta$ is not in the first quadrant and $\sin \theta$ is positive,$\theta$ must be in the second quadrant.
Therefore,$\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}$.
Now,$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \left(\frac{3}{5}\right) \left(-\frac{4}{5}\right) = -\frac{24}{25}$.
$\cos 2 \theta = 1 - 2 \sin^2 \theta = 1 - 2 \left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25}$.
$\tan 2 \theta = \frac{\sin 2 \theta}{\cos 2 \theta} = \frac{-24/25}{7/25} = -\frac{24}{7}$.
Substituting these values into the expression:
$15 \sin 2 \theta - 20 \cos 2 \theta - 7 \tan 2 \theta = 15 \left(-\frac{24}{25}\right) - 20 \left(\frac{7}{25}\right) - 7 \left(-\frac{24}{7}\right)$.
$= -\frac{72}{5} - \frac{28}{5} + 24 = -\frac{100}{5} + 24 = -20 + 24 = 4$.

(C) We know the half-angle formula: $\sin^2(\frac{A}{2}) = \frac{1 - \cos A}{2}$.
Let $A = 45^{\circ}$. Then $\frac{A}{2} = 22 \frac{1}{2}^{\circ}$.
Substituting the value of $A$:
$\sin^2(22 \frac{1}{2}^{\circ}) = \frac{1 - \cos 45^{\circ}}{2}$
$= \frac{1 - \frac{1}{\sqrt{2}}}{2}$
$= \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2} = \frac{\sqrt{2}-1}{2\sqrt{2}}$
To rationalize the denominator,multiply numerator and denominator by $\sqrt{2}$:
$= \frac{\sqrt{2}(\sqrt{2}-1)}{2\sqrt{2} \times \sqrt{2}} = \frac{2-\sqrt{2}}{4}$
Therefore,$\sin(22 \frac{1}{2}^{\circ}) = \sqrt{\frac{2-\sqrt{2}}{4}}$.

(A) We know that $\cos 2x = 2\cos^2 x - 1$,which implies $\cos^2 x = \frac{1 + \cos 2x}{2}$.
Now,$\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2$.
Expanding this,we get $\cos^4 x = \frac{1 + \cos^2 2x + 2\cos 2x}{4} = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{4}\cos^2 2x$.
Using the identity $\cos^2 2x = \frac{1 + \cos 4x}{2}$,we substitute:
$\cos^4 x = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{4}\left(\frac{1 + \cos 4x}{2}\right)$.
$\cos^4 x = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{8} + \frac{1}{8}\cos 4x$.
$\cos^4 x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.

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

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

(B) Let $A = \tan \left(\frac{7 \pi}{8}\right)$.
Since $\frac{7 \pi}{8} = \pi - \frac{\pi}{8}$,we have $A = \tan \left(\pi - \frac{\pi}{8}\right) = -\tan \left(\frac{\pi}{8}\right)$.
Using the formula $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,let $\theta = \frac{\pi}{8}$.
Then $\tan \left(\frac{\pi}{4}\right) = \frac{2 \tan \left(\frac{\pi}{8}\right)}{1 - \tan^2 \left(\frac{\pi}{8}\right)} = 1$.
Substituting $\tan \left(\frac{\pi}{8}\right) = -A$,we get $\frac{2(-A)}{1 - (-A)^2} = 1$.
$-2A = 1 - A^2 \Rightarrow A^2 - 2A - 1 = 0$.
Using the quadratic formula $A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $A = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.
Since $\frac{7 \pi}{8}$ lies in the $2^{nd}$ quadrant,$\tan \left(\frac{7 \pi}{8}\right) < 0$.
Thus,$A = 1 - \sqrt{2}$.

(C) We know that the formula for $\tan(2x)$ is $\frac{2 \tan(x)}{1-\tan^2(x)}$.
Given the inequality $\frac{2 \tan(x)}{1-\tan^2(x)} > 0$,we have $\tan(2x) > 0$.
Since $x \in \left(0, \frac{\pi}{2}\right)$,it follows that $2x \in (0, \pi)$.
For $\tan(2x) > 0$ in the interval $(0, \pi)$,$2x$ must lie in the first quadrant,i.e.,$0 < 2x < \frac{\pi}{2}$.
Dividing by $2$,we get $0 < x < \frac{\pi}{4}$.
Thus,the range of $x$ is $\left(0, \frac{\pi}{4}\right)$.

(D) We use the identity $\tan \theta - \cot \theta = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta} = -2 \cot 2 \theta$.
Given expression: $S = \tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha$.
Adding and subtracting $\cot \alpha$:
$S = (\tan \alpha - \cot \alpha) + \cot \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha$
$S = -2 \cot 2 \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$
$S = -2(\cot 2 \alpha - \tan 2 \alpha) + 4 \tan 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$
Since $\cot 2 \alpha - \tan 2 \alpha = 2 \cot 4 \alpha$,we have:
$S = -2(2 \cot 4 \alpha) + 4 \tan 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$
$S = -4 \cot 4 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha + \cot \alpha$
$S = -4(\cot 4 \alpha - \tan 4 \alpha) + 8 \cot 8 \alpha + \cot \alpha$
Since $\cot 4 \alpha - \tan 4 \alpha = 2 \cot 8 \alpha$,we have:
$S = -4(2 \cot 8 \alpha) + 8 \cot 8 \alpha + \cot \alpha$
$S = -8 \cot 8 \alpha + 8 \cot 8 \alpha + \cot \alpha = \cot \alpha$.

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

(A) We use the identity $\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta = \cos \theta (4 \cos^2 \theta - 3)$.
Thus,$4 \cos^2 \theta - 3 = \frac{\cos 3\theta}{\cos \theta}$.
Substituting this into the expression:
$(4 \cos^2 9^{\circ} - 3)(4 \cos^2 27^{\circ} - 3) = \left( \frac{\cos 27^{\circ}}{\cos 9^{\circ}} \right) \left( \frac{\cos 81^{\circ}}{\cos 27^{\circ}} \right)$
$= \frac{\cos 81^{\circ}}{\cos 9^{\circ}}$
$= \frac{\cos(90^{\circ} - 9^{\circ})}{\cos 9^{\circ}}$
$= \frac{\sin 9^{\circ}}{\cos 9^{\circ}} = \tan 9^{\circ}$.

(B) Let $A = 22 \frac{1}{2}^{\circ} = \frac{45^{\circ}}{2}$.
Using the half-angle formula $\cos A = \sqrt{\frac{1 + \cos 2A}{2}}$:
$\cos \left(\frac{45^{\circ}}{2}\right) = \sqrt{\frac{1 + \cos 45^{\circ}}{2}}$
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$\cos \left(22 \frac{1}{2}\right)^{\circ} = \sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}}$
$= \sqrt{\frac{\frac{\sqrt{2} + 1}{\sqrt{2}}}{2}}$
$= \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}}$
Thus,the correct option is $B$.

(B) We know that $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin(2\theta)}$.
Substituting $\theta = \frac{\pi}{24}$,we get $\tan \frac{\pi}{24} + \cot \frac{\pi}{24} = \frac{2}{\sin(\frac{\pi}{12})}$.
Since $\frac{\pi}{12} = 15^\circ$,$\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Thus,$\frac{1}{2} \left(\tan \frac{\pi}{24} + \cot \frac{\pi}{24}\right) = \frac{1}{2} \cdot \frac{2}{\sin(15^\circ)} = \frac{1}{\sin(15^\circ)} = \frac{2\sqrt{2}}{\sqrt{3}-1}$.
Rationalizing the denominator: $\frac{2\sqrt{2}(\sqrt{3}+1)}{3-1} = \sqrt{6} + \sqrt{2}$.
We are given $\sqrt{a^2+a} + \sqrt{a} = \sqrt{6} + \sqrt{2}$.
Comparing the terms,$\sqrt{a} = \sqrt{2} \implies a = 2$.
Checking the first term: $\sqrt{2^2+2} = \sqrt{6}$,which matches. Therefore,$a = 2$.

(B) Given,$\tan \frac{\theta}{2} = \operatorname{cosec} \theta - \sin \theta$
$\tan \frac{\theta}{2} = \frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta} = \frac{\cos^2 \theta}{\sin \theta}$
Using half-angle formulas $\cos \theta = \frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}$ and $\sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}$:
$\tan \frac{\theta}{2} = \frac{(\frac{1 - \tan^2 \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}})^2}{\frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}}} = \frac{(1 - \tan^2 \frac{\theta}{2})^2}{2 \tan \frac{\theta}{2} (1 + \tan^2 \frac{\theta}{2})}$
Let $x = \tan^2 \frac{\theta}{2}$. Then $\sqrt{x} = \frac{(1 - x)^2}{2 \sqrt{x} (1 + x)}$
$2x(1 + x) = (1 - x)^2$
$2x + 2x^2 = 1 - 2x + x^2$
$x^2 + 4x - 1 = 0$
Solving for $x$ using the quadratic formula: $x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$
Since $x = \tan^2 \frac{\theta}{2} > 0$,we have $x = -2 + \sqrt{5}$.

(D) We use the identity $\cos 3A = 4 \cos^3 A - 3 \cos A$,which implies $\cos^3 A = \frac{1}{4} (\cos 3A + 3 \cos A)$.
Applying this to each term:
$\cos^3 \theta = \frac{1}{4} (\cos 3\theta + 3 \cos \theta)$
$\cos^3(120^{\circ} + \theta) = \frac{1}{4} (\cos(360^{\circ} + 3\theta) + 3 \cos(120^{\circ} + \theta)) = \frac{1}{4} (\cos 3\theta + 3 \cos(120^{\circ} + \theta))$
$\cos^3(\theta - 120^{\circ}) = \frac{1}{4} (\cos(3\theta - 360^{\circ}) + 3 \cos(\theta - 120^{\circ})) = \frac{1}{4} (\cos 3\theta + 3 \cos(\theta - 120^{\circ}))$
Summing these up:
Sum $= \frac{1}{4} [3 \cos 3\theta + 3 (\cos \theta + \cos(120^{\circ} + \theta) + \cos(\theta - 120^{\circ}))]$
Using the sum-to-product formula $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$:
$\cos(120^{\circ} + \theta) + \cos(\theta - 120^{\circ}) = 2 \cos \theta \cos 120^{\circ} = 2 \cos \theta (-1/2) = -\cos \theta$
Substituting this back:
Sum $= \frac{1}{4} [3 \cos 3\theta + 3 (\cos \theta - \cos \theta)] = \frac{3}{4} \cos 3\theta$.

(C) Given equation: $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$
Expanding using $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:
$(\sin x \cos 3\alpha + \cos x \sin 3\alpha) + 3(\sin x \cos \alpha - \cos x \sin \alpha) = 0$
Using $\cos 3\alpha = 4\cos^3 \alpha - 3\cos \alpha$ and $\sin 3\alpha = 3\sin \alpha - 4\sin^3 \alpha$:
$\sin x(4\cos^3 \alpha - 3\cos \alpha) + \cos x(3\sin \alpha - 4\sin^3 \alpha) + 3\sin x \cos \alpha - 3\cos x \sin \alpha = 0$
$\sin x(4\cos^3 \alpha - 3\cos \alpha + 3\cos \alpha) + \cos x(3\sin \alpha - 4\sin^3 \alpha - 3\sin \alpha) = 0$
$4\sin x \cos^3 \alpha - 4\cos x \sin^3 \alpha = 0$
$4\sin x \cos^3 \alpha = 4\cos x \sin^3 \alpha$
$\frac{\sin x}{\cos x} = \frac{\sin^3 \alpha}{\cos^3 \alpha}$
$\tan x = \tan^3 \alpha$

(C) We are given that $\cosh(x) = \frac{5}{4}$.
Using the identity for hyperbolic functions,$\cosh(3x) = 4\cosh^3(x) - 3\cosh(x)$.
Substituting the given value:
$\cosh(3x) = 4\left(\frac{5}{4}\right)^3 - 3\left(\frac{5}{4}\right)$
$= 4\left(\frac{125}{64}\right) - \frac{15}{4}$
$= \frac{125}{16} - \frac{15}{4}$
$= \frac{125 - 60}{16} = \frac{65}{16}$.

(D) Given $540^{\circ} < A < 630^{\circ}$,which is $3\pi < A < \frac{7\pi}{2}$.
In this interval,$A$ lies in the third quadrant,so $\cos A < 0$ and $\tan A > 0$.
Since $|\cos A| = \frac{5}{13}$ and $\cos A < 0$,we have $\cos A = -\frac{5}{13}$.
Using $\tan A = \sqrt{\sec^2 A - 1}$,we get $\tan A = \sqrt{(\frac{13}{5})^2 - 1} = \sqrt{\frac{169}{25} - 1} = \sqrt{\frac{144}{25}} = \frac{12}{5}$.
For $\tan \frac{A}{2}$,we use the formula $\cos A = \frac{1 - \tan^2(A/2)}{1 + \tan^2(A/2)}$.
$-\frac{5}{13} = \frac{1 - \tan^2(A/2)}{1 + \tan^2(A/2)} \Rightarrow -5 - 5\tan^2(A/2) = 13 - 13\tan^2(A/2)$.
$8\tan^2(A/2) = 18 \Rightarrow \tan^2(A/2) = \frac{18}{8} = \frac{9}{4}$.
Since $540^{\circ} < A < 630^{\circ}$,then $270^{\circ} < \frac{A}{2} < 315^{\circ}$,which is the fourth quadrant.
In the fourth quadrant,$\tan \frac{A}{2} < 0$,so $\tan \frac{A}{2} = -\frac{3}{2}$.
Therefore,$\tan \frac{A}{2} \tan A = (-\frac{3}{2}) \times (\frac{12}{5}) = -\frac{18}{5}$.

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

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

(B) Given $\sin \theta = -\frac{3}{4}$.
We know that $\cos^2 \theta = 1 - \sin^2 \theta$.
$\cos^2 \theta = 1 - (-\frac{3}{4})^2 = 1 - \frac{9}{16} = \frac{7}{16}$.
Thus,$\cos \theta = \pm \frac{\sqrt{7}}{4}$.
Using the formula $\sin 2 \theta = 2 \sin \theta \cos \theta$:
If $\cos \theta = \frac{\sqrt{7}}{4}$,then $\sin 2 \theta = 2 \times (-\frac{3}{4}) \times (\frac{\sqrt{7}}{4}) = -\frac{3 \sqrt{7}}{8}$.
If $\cos \theta = -\frac{\sqrt{7}}{4}$,then $\sin 2 \theta = 2 \times (-\frac{3}{4}) \times (-\frac{\sqrt{7}}{4}) = \frac{3 \sqrt{7}}{8}$.
Since the options provided include $-\frac{3 \sqrt{7}}{8}$,the correct option is $B$.