Trigonometry Questions in English

(C) Given $b \sin \alpha = a \sin (\alpha + 2\beta)$.
$\frac{a}{b} = \frac{\sin \alpha}{\sin (\alpha + 2\beta)}$.
Using componendo and dividendo,we get:
$\frac{a + b}{a - b} = \frac{\sin \alpha + \sin (\alpha + 2\beta)}{\sin \alpha - \sin (\alpha + 2\beta)}$.
Applying the sum-to-product formulas $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:
Numerator: $\sin \alpha + \sin (\alpha + 2\beta) = 2 \sin(\alpha + \beta) \cos(-\beta) = 2 \sin(\alpha + \beta) \cos \beta$.
Denominator: $\sin \alpha - \sin (\alpha + 2\beta) = 2 \cos(\alpha + \beta) \sin(-\beta) = -2 \cos(\alpha + \beta) \sin \beta$.
Therefore,$\frac{a + b}{a - b} = \frac{2 \sin(\alpha + \beta) \cos \beta}{-2 \cos(\alpha + \beta) \sin \beta} = -\tan(\alpha + \beta) \cot \beta$.
This can be rewritten as $-\frac{\cot \beta}{\cot(\alpha + \beta)}$.

(B) Given expression: $\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}}$
Using the identity $\cos \theta = \sin(90^\circ - \theta)$:
$= \frac{{\sin (B + A) + \sin(90^\circ - (B - A))}}{{\sin (B - A) + \sin(90^\circ - (B + A))}}$
$= \frac{{\sin (B + A) + \sin(90^\circ - B + A)}}{{\sin (B - A) + \sin(90^\circ - B - A)}}$
Using the sum-to-product formula $\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})$:
Numerator: $2 \sin(\frac{B+A+90^\circ-B+A}{2}) \cos(\frac{B+A-90^\circ+B-A}{2}) = 2 \sin(45^\circ+A) \cos(B-45^\circ)$
Denominator: $2 \sin(\frac{B-A+90^\circ-B-A}{2}) \cos(\frac{B-A-90^\circ+B+A}{2}) = 2 \sin(45^\circ-A) \cos(B-45^\circ)$
Since $\cos(B-45^\circ) = \cos(45^\circ-B)$:
$= \frac{2 \sin(A+45^\circ) \cos(45^\circ-B)}{2 \sin(45^\circ-A) \cos(45^\circ-B)} = \frac{\sin(A+45^\circ)}{\sin(45^\circ-A)}$
Using $\sin(x+y) = \sin x \cos y + \cos x \sin y$ and $\sin(x-y) = \sin x \cos y - \cos x \sin y$:
$= \frac{\sin A \cos 45^\circ + \cos A \sin 45^\circ}{\sin 45^\circ \cos A - \cos 45^\circ \sin A}$
Dividing numerator and denominator by $\sin 45^\circ$ (or multiplying by $\sqrt{2}$):
$= \frac{\sin A + \cos A}{\cos A - \sin A} = \frac{\cos A + \sin A}{\cos A - \sin A}$.

(B) Given: $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b}$
Applying the Componendo and Dividendo rule,which states that if $\frac{p}{q} = \frac{r}{s},$ then $\frac{p + q}{p - q} = \frac{r + s}{r - s}$:
$\frac{\sin(x + y) + \sin(x - y)}{\sin(x + y) - \sin(x - y)} = \frac{(a + b) + (a - b)}{(a + b) - (a - b)}$
Using the trigonometric identities $\sin(A + B) + \sin(A - B) = 2\sin A \cos B$ and $\sin(A + B) - \sin(A - B) = 2\cos A \sin B$:
$\frac{2\sin x \cos y}{2\cos x \sin y} = \frac{2a}{2b}$
$\frac{\sin x}{\cos x} \cdot \frac{\cos y}{\sin y} = \frac{a}{b}$
$\tan x \cdot \frac{1}{\tan y} = \frac{a}{b}$
$\frac{\tan x}{\tan y} = \frac{a}{b}$

(A) Given: $x = \sin A + \sin 2A$ and $y = \cos A + \cos 2A$.
Calculate $x^2 + y^2$:
$x^2 + y^2 = (\sin A + \sin 2A)^2 + (\cos A + \cos 2A)^2$
$= \sin^2 A + \sin^2 2A + 2 \sin A \sin 2A + \cos^2 A + \cos^2 2A + 2 \cos A \cos 2A$
$= (\sin^2 A + \cos^2 A) + (\sin^2 2A + \cos^2 2A) + 2(\cos 2A \cos A + \sin 2A \sin A)$
$= 1 + 1 + 2 \cos(2A - A)$
$= 2 + 2 \cos A = 2(1 + \cos A)$.
Now,substitute this into the expression $({x^2} + {y^2})({x^2} + {y^2} - 3)$:
$= [2(1 + \cos A)][2(1 + \cos A) - 3]$
$= [2(1 + \cos A)][2 + 2 \cos A - 3]$
$= [2(1 + \cos A)][2 \cos A - 1]$
$= 2(2 \cos^2 A + \cos A - 1)$
$= 2(2 \cos^2 A - 1 + \cos A)$
$= 2(\cos 2A + \cos A) = 2y$.

(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$.
Using the identity $\cos nx = 2\cos((n-1)x)\cos x - \cos((n-2)x)$,or by expanding using binomial coefficients,we observe that the numerator is the real part of $(e^{ix} + e^{-ix})^6 / 2^6 \times 2^6$ or simply recognizing the expansion of $(2\cos x)^6$.
Alternatively,using the identity $\cos(nx) = \sum_{k=0}^{n} \binom{n}{k} \cos((n-2k)x)$,we can see that the numerator is $(2\cos x)^6 / 2 = 32\cos^6 x$.
However,a simpler way is to use the expansion of $(2\cos x)^n$.
We know that $(2\cos x)^6 = 2\cos 6x + 12\cos 4x + 30\cos 2x + 20$.
Dividing by $2$,we get $\cos 6x + 6\cos 4x + 15\cos 2x + 10 = \frac{(2\cos x)^6}{2} = 32\cos^6 x$.
Similarly,for the denominator $D = \cos 5x + 5\cos 3x + 10\cos x$,we know $(2\cos x)^5 = 2\cos 5x + 10\cos 3x + 20\cos x$.
Dividing by $2$,we get $\cos 5x + 5\cos 3x + 10\cos x = \frac{(2\cos x)^5}{2} = 16\cos^5 x$.
Therefore,the expression is $\frac{32\cos^6 x}{16\cos^5 x} = 2\cos x$.

(A) Expand each term using the identity $\sin(x - y) = \sin x \cos y - \cos x \sin y$.
$\cos \alpha (\sin \beta \cos \gamma - \cos \beta \sin \gamma) + \cos \beta (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) + \cos \gamma (\sin \alpha \cos \beta - \cos \alpha \sin \beta) = 0$.
Distributing the terms:
$\cos \alpha \sin \beta \cos \gamma - \cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha - \cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta - \cos \gamma \cos \alpha \sin \beta = 0$.
All terms cancel out:
$(\cos \alpha \sin \beta \cos \gamma - \cos \gamma \cos \alpha \sin \beta) + (-\cos \alpha \cos \beta \sin \gamma + \cos \beta \sin \gamma \cos \alpha) + (-\cos \beta \cos \gamma \sin \alpha + \cos \gamma \sin \alpha \cos \beta) = 0 + 0 + 0 = 0$.

(B) Let the expression be $E = \sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma )$.
Using the formula $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$:
First,combine the first two terms:
$\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) = 2 \sin \gamma \cos (\alpha - \beta) = 2 \sin \gamma \cos (\beta - \alpha)$.
Next,combine the last two terms:
$\sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma ) = 2 \cos (\alpha + \beta) \sin (-\gamma) = -2 \sin \gamma \cos (\alpha + \beta)$.
Now,combine these results:
$E = 2 \sin \gamma [\cos (\beta - \alpha) - \cos (\alpha + \beta)]$.
Using the identity $\cos (A-B) - \cos (A+B) = 2 \sin A \sin B$:
$E = 2 \sin \gamma [2 \sin \alpha \sin \beta] = 4 \sin \alpha \sin \beta \sin \gamma$.

(A) Given: $m \tan (\theta - 30^\circ) = n \tan (\theta + 120^\circ)$.
Rearranging the terms,we get: $\frac{m}{n} = \frac{\tan (\theta + 120^\circ)}{\tan (\theta - 30^\circ)}$.
Applying the Componendo and Dividendo rule:
$\frac{m + n}{m - n} = \frac{\tan (\theta + 120^\circ) + \tan (\theta - 30^\circ)}{\tan (\theta + 120^\circ) - \tan (\theta - 30^\circ)}$.
Using the identity $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$ and $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$,we get:
$\frac{m + n}{m - n} = \frac{\sin((\theta + 120^\circ) + (\theta - 30^\circ))}{\sin((\theta + 120^\circ) - (\theta - 30^\circ))}$.
$\frac{m + n}{m - n} = \frac{\sin(2\theta + 90^\circ)}{\sin(150^\circ)}$.
Since $\sin(2\theta + 90^\circ) = \cos 2\theta$ and $\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$.
Therefore,$\frac{m + n}{m - n} = \frac{\cos 2\theta}{1/2} = 2 \cos 2\theta$.

(A) Given expression: $2\cos x - (\cos 5x + \cos 3x)$
Using the sum-to-product formula $\cos C + \cos D = 2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$:
$\cos 5x + \cos 3x = 2\cos\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right) = 2\cos 4x \cos x$
Substituting this back into the expression:
$2\cos x - 2\cos 4x \cos x = 2\cos x(1 - \cos 4x)$
Using the identity $1 - \cos 2\theta = 2\sin^2 \theta$,where $2\theta = 4x$ (so $\theta = 2x$):
$2\cos x(2\sin^2 2x) = 4\cos x \sin^2 2x$
Using the double angle identity $\sin 2x = 2\sin x \cos x$:
$4\cos x (2\sin x \cos x)^2 = 4\cos x (4\sin^2 x \cos^2 x) = 16\sin^2 x \cos^3 x$.

(C) Given expression: $1 + \cos 2x + \cos 4x + \cos 6x$
Group the terms to use the sum-to-product formulas:
$= (1 + \cos 6x) + (\cos 2x + \cos 4x)$
Using the identities $1 + \cos 2\theta = 2\cos^2 \theta$ and $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$:
$= 2\cos^2 3x + 2\cos(\frac{4x+2x}{2})\cos(\frac{4x-2x}{2})$
$= 2\cos^2 3x + 2\cos 3x \cos x$
Factor out $2\cos 3x$:
$= 2\cos 3x (\cos 3x + \cos x)$
Apply the sum-to-product formula again for $(\cos 3x + \cos x)$:
$= 2\cos 3x [2\cos(\frac{3x+x}{2})\cos(\frac{3x-x}{2})]$
$= 2\cos 3x [2\cos 2x \cos x]$
$= 4\cos x \cos 2x \cos 3x$.

(A) Given the equation: $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$
Using the sum-to-product formulas:
$\sin A - \sin C = 2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}$
$\cos C - \cos A = 2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}$
Substituting these into the equation:
$\frac{2 \cos \frac{A + C}{2} \sin \frac{A - C}{2}}{2 \sin \frac{A + C}{2} \sin \frac{A - C}{2}} = \cot B$
$\cot \frac{A + C}{2} = \cot B$
This implies $\frac{A + C}{2} = B$,or $A + C = 2B$.
This is the condition for $A, B, C$ to be in $A.P.$ (Arithmetic Progression).

(D) We use the formula $\prod_{k=0}^{n-1} \cos(2^k \theta) = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Here,$\theta = \frac{2\pi}{15}$ and $n = 4$.
Therefore,$\cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{\sin(2^4 \cdot \frac{2\pi}{15})}{2^4 \sin \frac{2\pi}{15}}$.
$= \frac{\sin(\frac{32\pi}{15})}{16 \sin \frac{2\pi}{15}}$.
Since $\frac{32\pi}{15} = 2\pi + \frac{2\pi}{15}$,we have $\sin(\frac{32\pi}{15}) = \sin(2\pi + \frac{2\pi}{15}) = \sin \frac{2\pi}{15}$.
Substituting this back,we get $\frac{\sin \frac{2\pi}{15}}{16 \sin \frac{2\pi}{15}} = \frac{1}{16}$.

(A) Given expression: $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$
We know that $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$.
Using this identity:
$= \frac{1 + \cos(\frac{\pi}{6})}{2} + \cos^2(\frac{\pi}{4}) + \frac{1 + \cos(\frac{5\pi}{6})}{2}$
$= \frac{1}{2} + \frac{1}{2}\cos(\frac{\pi}{6}) + (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} + \frac{1}{2}\cos(\frac{5\pi}{6})$
$= 1 + \frac{1}{2} + \frac{1}{2}(\cos \frac{\pi}{6} + \cos \frac{5\pi}{6})$
Since $\cos(\frac{5\pi}{6}) = -\cos(\frac{\pi}{6})$:
$= \frac{3}{2} + \frac{1}{2}(\cos \frac{\pi}{6} - \cos \frac{\pi}{6})$
$= \frac{3}{2} + 0 = \frac{3}{2}$.

(B) We have the expression $E = \sin \frac{\pi }{16} \sin \frac{3\pi }{16} \sin \frac{5\pi }{16} \sin \frac{7\pi }{16}$.
Using the identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we can group the terms:
$E = \frac{1}{4} [ (2 \sin \frac{\pi }{16} \sin \frac{3\pi }{16}) (2 \sin \frac{5\pi }{16} \sin \frac{7\pi }{16}) ]$
$E = \frac{1}{4} [ (\cos \frac{2\pi }{16} - \cos \frac{4\pi }{16}) (\cos \frac{2\pi }{16} - \cos \frac{12\pi }{16}) ]$
$E = \frac{1}{4} [ (\cos \frac{\pi }{8} - \cos \frac{\pi }{4}) (\cos \frac{\pi }{8} - \cos \frac{3\pi }{4}) ]$
Since $\cos \frac{3\pi }{4} = -\cos \frac{\pi }{4} = -\frac{1}{\sqrt{2}}$,we have:
$E = \frac{1}{4} [ (\cos \frac{\pi }{8} - \frac{1}{\sqrt{2}}) (\cos \frac{\pi }{8} + \frac{1}{\sqrt{2}}) ]$
$E = \frac{1}{4} [ \cos^2 \frac{\pi }{8} - \frac{1}{2} ] = \frac{1}{8} [ 2 \cos^2 \frac{\pi }{8} - 1 ]$
Using the identity $\cos 2\theta = 2 \cos^2 \theta - 1$,we get:
$E = \frac{1}{8} [ \cos(2 \times \frac{\pi }{8}) ] = \frac{1}{8} \cos \frac{\pi }{4} = \frac{1}{8} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{16}$.

(D) We use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and the product-to-sum formula $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.
Given expression: $E = \cos^2 76^\circ + \cos^2 16^\circ - \cos 76^\circ \cos 16^\circ$
Multiply by $2/2$ to simplify:
$E = \frac{1}{2} [2\cos^2 76^\circ + 2\cos^2 16^\circ - 2\cos 76^\circ \cos 16^\circ]$
Using $2\cos^2 \theta = 1 + \cos 2\theta$:
$E = \frac{1}{2} [(1 + \cos 152^\circ) + (1 + \cos 32^\circ) - (\cos(76^\circ + 16^\circ) + \cos(76^\circ - 16^\circ))]$
$E = \frac{1}{2} [2 + \cos 152^\circ + \cos 32^\circ - \cos 92^\circ - \cos 60^\circ]$
Since $\cos 152^\circ = \cos(180^\circ - 28^\circ) = -\cos 28^\circ$ and $\cos 60^\circ = 1/2$:
$E = \frac{1}{2} [2 - \cos 28^\circ + \cos 32^\circ - \cos 92^\circ - 1/2]$
Alternatively,using $\cos 152^\circ + \cos 32^\circ = 2\cos(\frac{152+32}{2})\cos(\frac{152-32}{2}) = 2\cos 92^\circ \cos 60^\circ = 2\cos 92^\circ (1/2) = \cos 92^\circ$.
Substituting this back:
$E = \frac{1}{2} [2 + \cos 92^\circ - \cos 92^\circ - 1/2]$
$E = \frac{1}{2} [3/2] = 3/4$.

(D) We use the formula for the product of cosines: $\cos \theta \cos 2\theta \cos 4\theta \dots \cos 2^{n-1}\theta = \frac{\sin(2^n \theta)}{2^n \sin \theta}$.
Here,$\theta = \frac{\pi}{7}$ and $n = 3$.
Thus,$\cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7} = \frac{\sin(2^3 \cdot \frac{\pi}{7})}{2^3 \sin(\frac{\pi}{7})}$.
$= \frac{\sin(\frac{8\pi}{7})}{8 \sin(\frac{\pi}{7})}$.
Since $\sin(\frac{8\pi}{7}) = \sin(\pi + \frac{\pi}{7}) = -\sin(\frac{\pi}{7})$.
$= \frac{-\sin(\frac{\pi}{7})}{8 \sin(\frac{\pi}{7})} = -\frac{1}{8}$.

(B) We are given the expression $\frac{\tan 70^\circ - \tan 20^\circ}{\tan 50^\circ}$.
Using the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we have:
$\frac{\frac{\sin 70^\circ}{\cos 70^\circ} - \frac{\sin 20^\circ}{\cos 20^\circ}}{\frac{\sin 50^\circ}{\cos 50^\circ}}$
Simplifying the numerator:
$= \frac{\frac{\sin 70^\circ \cos 20^\circ - \cos 70^\circ \sin 20^\circ}{\cos 70^\circ \cos 20^\circ}}{\frac{\sin 50^\circ}{\cos 50^\circ}}$
Using the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{\sin(70^\circ - 20^\circ)}{\cos 70^\circ \cos 20^\circ} \times \frac{\cos 50^\circ}{\sin 50^\circ}$
$= \frac{\sin 50^\circ}{\cos 70^\circ \cos 20^\circ} \times \frac{\cos 50^\circ}{\sin 50^\circ} = \frac{\cos 50^\circ}{\cos 70^\circ \cos 20^\circ}$
Multiply numerator and denominator by $2$:
$= \frac{2 \cos 50^\circ}{2 \cos 70^\circ \cos 20^\circ}$
Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$:
$= \frac{2 \cos 50^\circ}{\cos(70^\circ + 20^\circ) + \cos(70^\circ - 20^\circ)}$
$= \frac{2 \cos 50^\circ}{\cos 90^\circ + \cos 50^\circ}$
Since $\cos 90^\circ = 0$:
$= \frac{2 \cos 50^\circ}{0 + \cos 50^\circ} = 2$.

(A) Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:
$\cos^2 \alpha + \cos^2(\alpha + 120^\circ) + \cos^2(\alpha - 120^\circ) = \frac{1 + \cos 2\alpha}{2} + \frac{1 + \cos(2\alpha + 240^\circ)}{2} + \frac{1 + \cos(2\alpha - 240^\circ)}{2}$
$= \frac{1}{2} [3 + \cos 2\alpha + \cos(2\alpha + 240^\circ) + \cos(2\alpha - 240^\circ)]$
Using $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$:
$= \frac{1}{2} [3 + \cos 2\alpha + 2\cos 2\alpha \cos 240^\circ]$
Since $\cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -1/2$:
$= \frac{1}{2} [3 + \cos 2\alpha + 2\cos 2\alpha (-1/2)]$
$= \frac{1}{2} [3 + \cos 2\alpha - \cos 2\alpha]$
$= \frac{3}{2}$.

(B) Given expression: $\tan 20^\circ + 2\tan 50^\circ - \tan 70^\circ$
$= (\tan 20^\circ - \tan 70^\circ) + 2\tan 50^\circ$
$= (\frac{\sin 20^\circ}{\cos 20^\circ} - \frac{\sin 70^\circ}{\cos 70^\circ}) + 2\tan 50^\circ$
$= \frac{\sin 20^\circ \cos 70^\circ - \cos 20^\circ \sin 70^\circ}{\cos 20^\circ \cos 70^\circ} + 2\tan 50^\circ$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$= \frac{\sin(20^\circ - 70^\circ)}{\cos 20^\circ \cos 70^\circ} + 2\tan 50^\circ$
$= \frac{\sin(-50^\circ)}{\cos 20^\circ \sin 20^\circ} + 2\tan 50^\circ$ (since $\cos 70^\circ = \sin 20^\circ$)
$= \frac{-2\sin 50^\circ}{2\sin 20^\circ \cos 20^\circ} + 2\tan 50^\circ$
$= \frac{-2\sin 50^\circ}{\sin 40^\circ} + 2\tan 50^\circ$
Since $\sin 40^\circ = \cos 50^\circ$:
$= \frac{-2\sin 50^\circ}{\cos 50^\circ} + 2\tan 50^\circ$
$= -2\tan 50^\circ + 2\tan 50^\circ = 0$.

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

(B) Given $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5}$.
Since $\theta$ and $\phi$ are acute angles,$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$ and $\sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$.
Using the formula $\cos(\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi$,we get:
$\cos(\theta - \phi) = (\frac{3}{5})(\frac{4}{5}) + (\frac{4}{5})(\frac{3}{5}) = \frac{12}{25} + \frac{12}{25} = \frac{24}{25}$.
Using the half-angle identity $2\cos^2(\frac{\theta - \phi}{2}) = 1 + \cos(\theta - \phi)$:
$2\cos^2(\frac{\theta - \phi}{2}) = 1 + \frac{24}{25} = \frac{49}{25}$.
$\cos^2(\frac{\theta - \phi}{2}) = \frac{49}{50}$.
Taking the square root,$\cos(\frac{\theta - \phi}{2}) = \sqrt{\frac{49}{50}} = \frac{7}{5\sqrt{2}}$.

(A) Given that $\sec \theta = 1\frac{1}{4} = \frac{5}{4}$.
We know the identity $\sec \theta = \frac{1 + \tan^2(\theta/2)}{1 - \tan^2(\theta/2)}$.
Substituting the value of $\sec \theta$:
$\frac{5}{4} = \frac{1 + \tan^2(\theta/2)}{1 - \tan^2(\theta/2)}$.
Cross-multiplying gives:
$5(1 - \tan^2(\theta/2)) = 4(1 + \tan^2(\theta/2))$.
$5 - 5\tan^2(\theta/2) = 4 + 4\tan^2(\theta/2)$.
$5 - 4 = 4\tan^2(\theta/2) + 5\tan^2(\theta/2)$.
$1 = 9\tan^2(\theta/2)$.
$\tan^2(\theta/2) = \frac{1}{9}$.
Taking the square root,$\tan(\theta/2) = \frac{1}{3}$.

(D) Given that $\tan \frac{A}{2} = \frac{3}{2}$.
We know the trigonometric identities: $1 + \cos A = 2 \cos^2 \frac{A}{2}$ and $1 - \cos A = 2 \sin^2 \frac{A}{2}$.
Substituting these into the expression:
$\frac{1 + \cos A}{1 - \cos A} = \frac{2 \cos^2 \frac{A}{2}}{2 \sin^2 \frac{A}{2}}$
$= \cot^2 \frac{A}{2}$
Since $\tan \frac{A}{2} = \frac{3}{2}$,then $\cot \frac{A}{2} = \frac{1}{\tan \frac{A}{2}} = \frac{2}{3}$.
Therefore,$\cot^2 \frac{A}{2} = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.

(D) Given that $\cos A = \frac{\sqrt{3}}{2}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $A = 30^{\circ}$.
Now,we need to find the value of $\tan 3A$.
Substituting $A = 30^{\circ}$,we get $\tan 3(30^{\circ}) = \tan 90^{\circ}$.
Since $\tan 90^{\circ}$ is undefined,the value is $\infty$.

(A) We know that $\sin 2A = 2\sin A \cos A$.
Applying this to $\sin 4\theta$,we get $\sin 4\theta = 2\sin 2\theta \cos 2\theta$.
Using the identities $\sin 2\theta = 2\sin \theta \cos \theta$ and $\cos 2\theta = 1 - 2\sin^2 \theta$,we substitute these into the expression:
$\sin 4\theta = 2(2\sin \theta \cos \theta)(1 - 2\sin^2 \theta)$.
Since $\cos \theta = \sqrt{1 - \sin^2 \theta}$,we substitute this to express the entire equation in terms of $\sin \theta$:
$\sin 4\theta = 4\sin \theta \sqrt{1 - \sin^2 \theta} (1 - 2\sin^2 \theta)$.

(B) Given: $\cos 2B = \frac{\cos (A + C)}{\cos (A - C)}$
Using the expansion formulas for $\cos(A+C)$ and $\cos(A-C)$:
$\cos 2B = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$
Divide the numerator and denominator by $\cos A \cos C$:
$\cos 2B = \frac{1 - \tan A \tan C}{1 + \tan A \tan C}$
We know that $\cos 2B = \frac{1 - \tan^2 B}{1 + \tan^2 B}$.
Equating the two expressions:
$\frac{1 - \tan^2 B}{1 + \tan^2 B} = \frac{1 - \tan A \tan C}{1 + \tan A \tan C}$
By applying componendo and dividendo or cross-multiplying:
$(1 - \tan^2 B)(1 + \tan A \tan C) = (1 + \tan^2 B)(1 - \tan A \tan C)$
$1 + \tan A \tan C - \tan^2 B - \tan^2 B \tan A \tan C = 1 - \tan A \tan C + \tan^2 B - \tan^2 B \tan A \tan C$
Simplifying the equation:
$2 \tan A \tan C = 2 \tan^2 B$
$\tan^2 B = \tan A \tan C$
Since the square of the middle term is equal to the product of the first and third terms,$\tan A, \tan B, \tan C$ are in $G.P.$

(A) Given that $a \tan \theta = b$,which implies $\tan \theta = \frac{b}{a}$.
We need to evaluate the expression $a \cos 2\theta + b \sin 2\theta$.
Using the double angle formulas in terms of $\tan \theta$:
$\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$.
Substituting these into the expression:
$a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)$
Substitute $\tan \theta = \frac{b}{a}$:
$= a \left( \frac{1 - \frac{b^2}{a^2}}{1 + \frac{b^2}{a^2}} \right) + b \left( \frac{2 \frac{b}{a}}{1 + \frac{b^2}{a^2}} \right)$
$= a \left( \frac{\frac{a^2 - b^2}{a^2}}{\frac{a^2 + b^2}{a^2}} \right) + b \left( \frac{\frac{2b}{a}}{\frac{a^2 + b^2}{a^2}} \right)$
$= a \left( \frac{a^2 - b^2}{a^2 + b^2} \right) + b \left( \frac{2ab}{a^2 + b^2} \right)$
$= \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2}$
$= \frac{a^3 + ab^2}{a^2 + b^2} = \frac{a(a^2 + b^2)}{a^2 + b^2} = a$.

(A) Given expression: $\left( \frac{\sin 2A}{1 + \cos 2A} \right) \left( \frac{\cos A}{1 + \cos A} \right)$
Using trigonometric identities:
$\sin 2A = 2 \sin A \cos A$
$1 + \cos 2A = 2 \cos^2 A$
Substituting these into the first part:
$\frac{2 \sin A \cos A}{2 \cos^2 A} = \frac{\sin A}{\cos A} = \tan A$
Now,the expression becomes:
$\tan A \cdot \frac{\cos A}{1 + \cos A} = \frac{\sin A}{\cos A} \cdot \frac{\cos A}{1 + \cos A} = \frac{\sin A}{1 + \cos A}$
Using half-angle formulas:
$\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$
$1 + \cos A = 2 \cos^2 \frac{A}{2}$
Therefore:
$\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \cos^2 \frac{A}{2}} = \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} = \tan \frac{A}{2}$.

(D) Given expression: $\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}}$
We know that $\cot \theta = \frac{1}{\tan \theta}$. Substituting this into the second term:
$\frac{1}{{\cot 3A - \cot A}} = \frac{1}{{\frac{1}{\tan 3A} - \frac{1}{\tan A}}} = \frac{1}{{\frac{\tan A - \tan 3A}{\tan 3A \tan A}}} = \frac{\tan 3A \tan A}{\tan A - \tan 3A} = -\frac{\tan 3A \tan A}{\tan 3A - \tan A}$
Substituting this back into the original expression:
$= \frac{1}{{\tan 3A - \tan A}} - \left( -\frac{\tan 3A \tan A}{\tan 3A - \tan A} \right)$
$= \frac{1 + \tan 3A \tan A}{\tan 3A - \tan A}$
Using the identity $\tan(X - Y) = \frac{\tan X - \tan Y}{1 + \tan X \tan Y}$,we have $\tan(3A - A) = \frac{\tan 3A - \tan A}{1 + \tan 3A \tan A}$.
Therefore,$\frac{1 + \tan 3A \tan A}{\tan 3A - \tan A} = \frac{1}{\tan(3A - A)} = \frac{1}{\tan 2A} = \cot 2A$.

(A) Given expression: $\text{cosec } A - 2 \cot 2A \cos A$
$= \frac{1}{\sin A} - \frac{2 \cos A \cos 2A}{\sin 2A}$
$= \frac{1}{\sin A} - \frac{2 \cos A \cos 2A}{2 \sin A \cos A}$
$= \frac{1}{\sin A} - \frac{\cos 2A}{\sin A}$
$= \frac{1 - \cos 2A}{\sin A}$
Using the identity $1 - \cos 2A = 2 \sin^2 A$:
$= \frac{2 \sin^2 A}{\sin A}$
$= 2 \sin A$.

(A) Given expression: $\sqrt {2 + \sqrt {2 + 2\cos 4\theta } }$
Using the trigonometric identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 4\theta = 2(1 + \cos 4\theta) = 2(2\cos^2 2\theta) = 4\cos^2 2\theta$.
Substituting this into the inner square root:
$\sqrt {2 + \sqrt {4\cos^2 2\theta } } = \sqrt {2 + 2\cos 2\theta }$
Again,using the identity $1 + \cos 2A = 2\cos^2 A$,we have $2 + 2\cos 2\theta = 2(1 + \cos 2\theta) = 2(2\cos^2 \theta) = 4\cos^2 \theta$.
Taking the square root:
$\sqrt {4\cos^2 \theta } = 2\cos \theta$.
Thus,the correct option is $A$.

(A) Expand the squares:
$(\cos ^2 \alpha + \cos ^2 \beta + 2\cos \alpha \cos \beta) + (\sin ^2 \alpha + \sin ^2 \beta + 2\sin \alpha \sin \beta)$
Group the terms using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2(\cos \alpha \cos \beta + \sin \alpha \sin \beta)$
$= 1 + 1 + 2\cos(\alpha - \beta)$
$= 2 + 2\cos(\alpha - \beta)$
$= 2(1 + \cos(\alpha - \beta))$
Using the identity $1 + \cos(2\theta) = 2\cos^2 \theta$,we get:
$= 2(2\cos^2(\frac{\alpha - \beta}{2}))$
$= 4\cos^2(\frac{\alpha - \beta}{2})$.

(B) Given that,$\tan x = \frac{b}{a}.$
Let the expression be $E = \sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}.$
Dividing the numerator and denominator of each term inside the square roots by $a,$ we get:
$E = \sqrt {\frac{{1 + b/a}}{{1 - b/a}}} + \sqrt {\frac{{1 - b/a}}{{1 + b/a}}} = \sqrt {\frac{{1 + \tan x}}{{1 - \tan x}}} + \sqrt {\frac{{1 - \tan x}}{{1 + \tan x}}}.$
Taking the common denominator:
$E = \frac{{(1 + \tan x) + (1 - \tan x)}}{{\sqrt {(1 - \tan x)(1 + \tan x)} }} = \frac{2}{{\sqrt {1 - \tan^2 x} }}.$
Substituting $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}:$
$E = \frac{2}{{\sqrt {1 - \frac{\sin^2 x}{\cos^2 x}} }} = \frac{2}{{\sqrt {\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} }} = \frac{2\cos x}{{\sqrt {\cos 2x} }}.$

(D) Given expression: $\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}}$
Using trigonometric identities:
$1$. $\cos \left( {\frac{\pi }{2} - A} \right) = \sin A$
$2$. $\cos (\pi + 3A) = -\cos 3A$
Substituting these into the expression:
$= \frac{{\sin 3A - \sin A}}{{\cos A - \cos 3A}}$
Applying sum-to-product formulas:
$\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$
$\cos D - \cos C = 2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$
Numerator: $\sin 3A - \sin A = 2 \cos 2A \sin A$
Denominator: $\cos A - \cos 3A = 2 \sin 2A \sin A$
$= \frac{{2 \cos 2A \sin A}}{{2 \sin 2A \sin A}}$
$= \frac{{\cos 2A}}{{\sin 2A}} = \cot 2A$.

(B) We know the formula for $\tan 3A$ is given by:
$\tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$
Given $\tan A = \frac{1}{2}$.
Substituting the value of $\tan A$ in the formula:
$\tan 3A = \frac{3(\frac{1}{2}) - (\frac{1}{2})^3}{1 - 3(\frac{1}{2})^2}$
$= \frac{\frac{3}{2} - \frac{1}{8}}{1 - 3(\frac{1}{4})}$
$= \frac{\frac{12 - 1}{8}}{1 - \frac{3}{4}}$
$= \frac{\frac{11}{8}}{\frac{4 - 3}{4}}$
$= \frac{11}{8} \times \frac{4}{1} = \frac{11}{2}$.

(B) Given expression: $E = \frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}$
We know that $1 + \sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$
And $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$
Since $x$ lies in the $II^{nd}$ quadrant,$\frac{\pi}{2} < x < \pi$,which implies $\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$.
In this interval,$\cos \frac{x}{2} > 0$ and $\sin \frac{x}{2} > 0$,and $\cos \frac{x}{2} > \sin \frac{x}{2}$.
Thus,$\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$.
Substituting these into the expression:
$E = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}$
$E = \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} = \cot \frac{x}{2}$.

(D) Given expression: $(\sec 2A + 1){\sec ^2}A$
We know that $\sec 2A = \frac{1}{\cos 2A} = \frac{1 + \tan^2 A}{1 - \tan^2 A}$ and $\sec^2 A = 1 + \tan^2 A$.
Substituting these into the expression:
$= \left( \frac{1 + \tan^2 A}{1 - \tan^2 A} + 1 \right) (1 + \tan^2 A)$
$= \left( \frac{1 + \tan^2 A + 1 - \tan^2 A}{1 - \tan^2 A} \right) (1 + \tan^2 A)$
$= \left( \frac{2}{1 - \tan^2 A} \right) (1 + \tan^2 A)$
$= 2 \cdot \frac{1 + \tan^2 A}{1 - \tan^2 A}$
Since $\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A}$,it follows that $\sec 2A = \frac{1 + \tan^2 A}{1 - \tan^2 A}$.
Therefore,the expression becomes $2 \sec 2A$.

(B) Given expression: $2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A$
Step $1$: Factor out $2\sin A\cos A$ from the expression:
$= 2\sin A\cos A(\cos^2 A - \sin^2 A)$
Step $2$: Use the double angle identities $\sin 2A = 2\sin A\cos A$ and $\cos 2A = \cos^2 A - \sin^2 A$:
$= \sin 2A \cdot \cos 2A$
Step $3$: Multiply and divide by $2$ to use the identity $\sin 2\theta = 2\sin \theta \cos \theta$ where $\theta = 2A$:
$= \frac{1}{2}(2\sin 2A \cos 2A)$
$= \frac{1}{2}\sin 4A$.

(C) Given expression: $\frac{\sin \theta + \sin 2\theta }{1 + \cos \theta + \cos 2\theta }$
Using trigonometric identities: $\sin 2\theta = 2\sin \theta \cos \theta$ and $\cos 2\theta = 2\cos^2 \theta - 1$.
Substitute these into the expression:
$= \frac{\sin \theta + 2\sin \theta \cos \theta }{1 + \cos \theta + (2\cos^2 \theta - 1)}$
$= \frac{\sin \theta (1 + 2\cos \theta )}{\cos \theta + 2\cos^2 \theta }$
$= \frac{\sin \theta (1 + 2\cos \theta )}{\cos \theta (1 + 2\cos \theta )}$
$= \frac{\sin \theta }{\cos \theta } = \tan \theta $.
Thus,the correct option is $C$.

(B) Given,$\frac{2\sin \alpha}{1 + \cos \alpha + \sin \alpha} = y$.
Using half-angle formulas,$\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$,$1 + \cos \alpha = 2\cos^2 \frac{\alpha}{2}$.
Substituting these,we get $\frac{2(2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2})}{2\cos^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = y$.
Dividing numerator and denominator by $2\cos \frac{\alpha}{2}$,we get $\frac{2\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2} + \sin \frac{\alpha}{2}} = y$.
Now,consider the expression $E = \frac{1 - \cos \alpha + \sin \alpha}{1 + \sin \alpha}$.
Using $1 - \cos \alpha = 2\sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$,the numerator becomes $2\sin^2 \frac{\alpha}{2} + 2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} = 2\sin \frac{\alpha}{2} (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})$.
The denominator $1 + \sin \alpha = (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2$.
Thus,$E = \frac{2\sin \frac{\alpha}{2} (\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})}{(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})^2} = \frac{2\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2}} = y$.

(A) Given $\sin \beta = \frac{1}{\sqrt{10}}$. Since $0 < \beta < \frac{\pi}{2}$,we have $\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Thus,$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$.
Now,calculate $\tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$.
Next,consider $\tan(\alpha + 2\beta) = \frac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha \tan 2\beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{(4+21)/28}{(28-3)/28} = \frac{25}{25} = 1$.
Since $0 < \alpha < \frac{\pi}{2}$ and $0 < \beta < \frac{\pi}{2}$,we have $0 < 2\beta < \frac{\pi}{2}$ (as $\tan 2\beta = 3/4 > 0$).
Thus,$0 < \alpha + 2\beta < \pi$. Since $\tan(\alpha + 2\beta) = 1$,we must have $\alpha + 2\beta = \frac{\pi}{4}$.
Therefore,$2\beta = \frac{\pi}{4} - \alpha$.

(A) Given that $\cos (\theta - \alpha ), \cos \theta, \cos (\theta + \alpha )$ are in $H.P.$
Therefore,their reciprocals $\frac{1}{\cos (\theta - \alpha )}, \frac{1}{\cos \theta}, \frac{1}{\cos (\theta + \alpha )}$ are in $A.P.$
This implies: $\frac{2}{\cos \theta} = \frac{1}{\cos (\theta - \alpha )} + \frac{1}{\cos (\theta + \alpha )}$
$\frac{2}{\cos \theta} = \frac{\cos (\theta + \alpha ) + \cos (\theta - \alpha )}{\cos (\theta - \alpha ) \cos (\theta + \alpha )}$
Using the identity $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$ and $\cos (A-B) \cos (A+B) = \cos^2 A - \sin^2 B$:
$\frac{2}{\cos \theta} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$
$\cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha$
$\cos^2 \theta (1 - \cos \alpha ) = \sin^2 \alpha$
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin \alpha = 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}$:
$\cos^2 \theta (2 \sin^2 \frac{\alpha}{2}) = (2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2})^2$
$\cos^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$
$\cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$
$\cos^2 \theta \sec^2 \frac{\alpha}{2} = 2$
Taking the square root on both sides:
$\cos \theta \sec \frac{\alpha}{2} = \pm \sqrt{2}$.

(B) Given: $\sin \theta + \sin \phi = a$ $(i)$ and $\cos \theta + \cos \phi = b$ $(ii)$.
Squaring and adding $(i)$ and $(ii)$:
$(\sin \theta + \sin \phi)^2 + (\cos \theta + \cos \phi)^2 = a^2 + b^2$
$(\sin^2 \theta + \cos^2 \theta) + (\sin^2 \phi + \cos^2 \phi) + 2(\sin \theta \sin \phi + \cos \theta \cos \phi) = a^2 + b^2$
$1 + 1 + 2 \cos(\theta - \phi) = a^2 + b^2$
$2 \cos(\theta - \phi) = a^2 + b^2 - 2$
$\cos(\theta - \phi) = \frac{a^2 + b^2 - 2}{2}$.
Using the identity $\cos x = \frac{1 - \tan^2(x/2)}{1 + \tan^2(x/2)}$:
$\frac{1 - \tan^2(\frac{\theta - \phi}{2})}{1 + \tan^2(\frac{\theta - \phi}{2})} = \frac{a^2 + b^2 - 2}{2}$.
Applying componendo and dividendo:
$\frac{(1 + \tan^2(\frac{\theta - \phi}{2})) + (1 - \tan^2(\frac{\theta - \phi}{2}))}{(1 + \tan^2(\frac{\theta - \phi}{2})) - (1 - \tan^2(\frac{\theta - \phi}{2}))} = \frac{2 + (a^2 + b^2 - 2)}{2 - (a^2 + b^2 - 2)}$
$\frac{2}{2 \tan^2(\frac{\theta - \phi}{2})} = \frac{a^2 + b^2}{4 - a^2 - b^2}$
$\tan^2(\frac{\theta - \phi}{2}) = \frac{4 - a^2 - b^2}{a^2 + b^2}$
$\tan(\frac{\theta - \phi}{2}) = \sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}}$.

(A) Given $\tan A = \frac{1 - \cos B}{\sin B}$.
Using the half-angle identities $1 - \cos B = 2\sin^2(B/2)$ and $\sin B = 2\sin(B/2)\cos(B/2)$:
$\tan A = \frac{2\sin^2(B/2)}{2\sin(B/2)\cos(B/2)} = \frac{\sin(B/2)}{\cos(B/2)} = \tan(B/2)$.
Therefore,$A = B/2$,which implies $2A = B$.
Thus,$\tan 2A = \tan B$.

(D) Since $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$
$\sin^2 \beta = \sin \alpha \cos \alpha.$
Now,$\cos 2\beta = 1 - 2\sin^2 \beta = 1 - 2\sin \alpha \cos \alpha.$
Using the identity $\sin 2\alpha = 2\sin \alpha \cos \alpha,$ we get $\cos 2\beta = 1 - \sin 2\alpha.$
Since $1 = \sin^2 \alpha + \cos^2 \alpha,$ we have $\cos 2\beta = \cos^2 \alpha + \sin^2 \alpha - 2\sin \alpha \cos \alpha = (\cos \alpha - \sin \alpha)^2.$
We can write this as $2\left( \frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alpha \right)^2 = 2\left( \sin \frac{\pi}{4} \cos \alpha - \cos \frac{\pi}{4} \sin \alpha \right)^2 = 2\sin^2\left( \frac{\pi}{4} - \alpha \right).$
This matches option $(a).$
Also,using $\cos^2 \theta = \sin^2(\frac{\pi}{2} - \theta),$ we have $2\sin^2\left( \frac{\pi}{4} - \alpha \right) = 2\cos^2\left( \frac{\pi}{2} - (\frac{\pi}{4} - \alpha) \right) = 2\cos^2\left( \frac{\pi}{4} + \alpha \right).$
This matches option $(b).$
Therefore,both $(a)$ and $(b)$ are correct.

(B) Given expression: $\frac{\sec 8A - 1}{\sec 4A - 1}$
Using $\sec \theta = \frac{1}{\cos \theta}$,we get:
$= \frac{\frac{1}{\cos 8A} - 1}{\frac{1}{\cos 4A} - 1} = \frac{1 - \cos 8A}{\cos 8A} \cdot \frac{\cos 4A}{1 - \cos 4A}$
Using the identity $1 - \cos 2\theta = 2\sin^2 \theta$:
$= \frac{2\sin^2 4A}{\cos 8A} \cdot \frac{\cos 4A}{2\sin^2 2A}$
$= \frac{(2\sin 4A \cos 4A) \cdot \sin 4A}{\cos 8A \cdot 2\sin^2 2A}$
Using $\sin 2\theta = 2\sin \theta \cos \theta$,so $\sin 8A = 2\sin 4A \cos 4A$:
$= \frac{\sin 8A}{\cos 8A} \cdot \frac{\sin 4A}{2\sin^2 2A}$
$= \tan 8A \cdot \frac{2\sin 2A \cos 2A}{2\sin^2 2A}$
$= \tan 8A \cdot \frac{\cos 2A}{\sin 2A} = \tan 8A \cdot \cot 2A = \frac{\tan 8A}{\tan 2A}$

(C) We use the product-to-sum formula: $2\sin x \sin y = \cos(x - y) - \cos(x + y)$.
Applying this to the expression $32\sin \left( \frac{A}{2} \right)\sin \left( \frac{5A}{2} \right)$:
$= 16 \times [2\sin \left( \frac{A}{2} \right)\sin \left( \frac{5A}{2} \right)]$
$= 16 \times [\cos \left( \frac{5A}{2} - \frac{A}{2} \right) - \cos \left( \frac{5A}{2} + \frac{A}{2} \right)]$
$= 16 \times [\cos(2A) - \cos(3A)]$
Using the identities $\cos(2A) = 2\cos^2 A - 1$ and $\cos(3A) = 4\cos^3 A - 3\cos A$:
$= 16 \times [(2\cos^2 A - 1) - (4\cos^3 A - 3\cos A)]$
Given $\cos A = \frac{3}{4}$,substitute the value:
$= 16 \times [2(\frac{3}{4})^2 - 1 - 4(\frac{3}{4})^3 + 3(\frac{3}{4})]$
$= 16 \times [2(\frac{9}{16}) - 1 - 4(\frac{27}{64}) + \frac{9}{4}]$
$= 16 \times [\frac{9}{8} - 1 - \frac{27}{16} + \frac{9}{4}]$
$= 16 \times [\frac{18 - 16 - 27 + 36}{16}]$
$= 18 - 16 - 27 + 36 = 11$.

(C) We know that $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
$\tan 15^\circ = \tan(45^\circ - 30^\circ)$
$= \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$
$= \frac{1 - 1/\sqrt{3}}{1 + 1 \cdot (1/\sqrt{3})} = \frac{(\sqrt{3} - 1)/\sqrt{3}}{(\sqrt{3} + 1)/\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Rationalizing the denominator:
$= \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

(B) Given $\tan \alpha = \frac{1}{7}$.
Using the formula $\cos 2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}$:
$\cos 2\alpha = \frac{1 - (1/7)^2}{1 + (1/7)^2} = \frac{1 - 1/49}{1 + 1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25}$.
Given $\tan \beta = \frac{1}{3}$.
Using the formula $\sin 2\beta = \frac{2 \tan \beta}{1 + \tan^2 \beta}$:
$\sin 2\beta = \frac{2(1/3)}{1 + (1/3)^2} = \frac{2/3}{1 + 1/9} = \frac{2/3}{10/9} = \frac{2}{3} \times \frac{9}{10} = \frac{3}{5}$.
Since $\sin 2\beta = \frac{3}{5}$,then $\cos 2\beta = \sqrt{1 - (3/5)^2} = \frac{4}{5}$.
Now,$\sin 4\beta = 2 \sin 2\beta \cos 2\beta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$.
Thus,$\cos 2\alpha = \sin 4\beta$.

(C) Given that $\tan \beta = \cos \theta \tan \alpha$,we have $\cos \theta = \frac{\tan \beta}{\tan \alpha}$.
Using the identity $\tan^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{1 + \cos \theta}$,we substitute $\cos \theta$:
$\tan^2 \frac{\theta}{2} = \frac{1 - \frac{\tan \beta}{\tan \alpha}}{1 + \frac{\tan \beta}{\tan \alpha}}$
$= \frac{\tan \alpha - \tan \beta}{\tan \alpha + \tan \beta}$
$= \frac{\frac{\sin \alpha}{\cos \alpha} - \frac{\sin \beta}{\cos \beta}}{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}$
$= \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\sin \alpha \cos \beta + \cos \alpha \sin \beta}$
$= \frac{\sin (\alpha - \beta)}{\sin (\alpha + \beta)}$.