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

(A) Using the sum-to-product formula $\cos C + \cos D = 2 \cos \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$:
$\cos 52^\circ + \cos 172^\circ + \cos 68^\circ$
$= 2 \cos \left( \frac{52^\circ + 172^\circ}{2} \right) \cos \left( \frac{52^\circ - 172^\circ}{2} \right) + \cos 68^\circ$
$= 2 \cos (112^\circ) \cos (-60^\circ) + \cos 68^\circ$
Since $\cos (-60^\circ) = \cos 60^\circ = \frac{1}{2}$:
$= 2 \cos (112^\circ) \left( \frac{1}{2} \right) + \cos 68^\circ$
$= \cos 112^\circ + \cos 68^\circ$
$= \cos (180^\circ - 68^\circ) + \cos 68^\circ$
$= -\cos 68^\circ + \cos 68^\circ = 0$.

(C) Given equations are $\cos x + \cos y = -\cos \alpha$ and $\sin x + \sin y = -\sin \alpha$.
Using the sum-to-product formulas,we have:
$2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right) = -\cos \alpha$ $(i)$
$2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right) = -\sin \alpha$ $(ii)$
Dividing equation $(ii)$ by equation $(i)$,we get:
$\frac{2 \sin \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)}{2 \cos \left( \frac{x + y}{2} \right) \cos \left( \frac{x - y}{2} \right)} = \frac{-\sin \alpha}{-\cos \alpha}$
$\tan \left( \frac{x + y}{2} \right) = \tan \alpha$
Taking the reciprocal on both sides:
$\cot \left( \frac{x + y}{2} \right) = \cot \alpha$.

(A) Given equations are:
$\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha$ ... $(i)$
$\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha$ ... $(ii)$
Using the sum-to-product formula $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ in $(i)$:
$(\sin 3\theta + \sin \theta) + \sin 2\theta = \sin \alpha$
$2 \sin 2\theta \cos \theta + \sin 2\theta = \sin \alpha$
$\sin 2\theta (2 \cos \theta + 1) = \sin \alpha$ ... $(iii)$
Using the sum-to-product formula $\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$ in $(ii)$:
$(\cos 3\theta + \cos \theta) + \cos 2\theta = \cos \alpha$
$2 \cos 2\theta \cos \theta + \cos 2\theta = \cos \alpha$
$\cos 2\theta (2 \cos \theta + 1) = \cos \alpha$ ... $(iv)$
Dividing $(iii)$ by $(iv)$:
$\frac{\sin 2\theta (2 \cos \theta + 1)}{\cos 2\theta (2 \cos \theta + 1)} = \frac{\sin \alpha}{\cos \alpha}$
$\tan 2\theta = \tan \alpha$
$2\theta = \alpha$
$\theta = \alpha / 2$

(C) We know that $\tan {50^o} = \tan ({70^o} - {20^o}) = \frac{\tan {70^o} - \tan {20^o}}{1 + \tan {70^o} \tan {20^o}}$.
Since $\tan {70^o} = \cot {20^o} = \frac{1}{\tan {20^o}}$,we have $\tan {70^o} \tan {20^o} = 1$.
Substituting this into the equation: $\tan {50^o} = \frac{\tan {70^o} - \tan {20^o}}{1 + 1} = \frac{\tan {70^o} - \tan {20^o}}{2}$.
Thus,$2 \tan {50^o} = \tan {70^o} - \tan {20^o}$,which implies $\tan {70^o} = 2 \tan {50^o} + \tan {20^o}$.
Also,$\sec {50^o} + \tan {50^o} = \frac{1 + \sin {50^o}}{\cos {50^o}} = \frac{\cos {40^o} + \sin {40^o}}{\sin {40^o}} = \cot {40^o} + 1 = \tan {50^o} + 1$ (This is not the direct path).
Using the identity $\tan {70^o} = \sec {50^o} + \tan {50^o}$:
Since $\tan {70^o} = \tan (90^o - 20^o) = \cot {20^o} = \frac{\cos {20^o}}{\sin {20^o}} = \frac{1 + \cos {40^o}}{\sin {40^o}} = \frac{2 \cos^2 {20^o}}{2 \sin {20^o} \cos {20^o}} = \cot {20^o}$.
Actually,$\sec {50^o} + \tan {50^o} = \tan (45^o + 25^o) = \tan {70^o} = 2 \tan {50^o} + \tan {20^o}$ is incorrect.
Correct derivation: $\sec {50^o} + \tan {50^o} = \tan (45^o + 25^o)$.
Given the options,the correct identity is $\sec {50^o} + \tan {50^o} = \tan {20^o} + 2 \tan {50^o}$.

(A) The sum of the series is given by $S = \sum_{k=1}^{n} \sin(k\theta)$.
Using the formula for the sum of sines in arithmetic progression: $\sum_{k=0}^{n-1} \sin(\alpha + k\beta) = \frac{\sin(n\beta/2)}{\sin(\beta/2)} \sin(\alpha + \frac{(n-1)\beta}{2})$.
Here,$\alpha = \theta$ and $\beta = \theta$.
Substituting these values,we get:
$S = \frac{\sin(n\theta/2)}{\sin(\theta/2)} \sin(\theta + \frac{(n-1)\theta}{2})$
$S = \frac{\sin(n\theta/2)}{\sin(\theta/2)} \sin(\frac{2\theta + n\theta - \theta}{2})$
$S = \frac{\sin(n\theta/2) \sin(\frac{(n+1)\theta}{2})}{\sin(\theta/2)}$.

(B) Let the expression be $E = 2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$.
Using the product-to-sum formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:
$2 \cos \frac{\pi}{13} \cos \frac{9\pi}{13} = \cos(\frac{10\pi}{13}) + \cos(-\frac{8\pi}{13}) = \cos(\frac{10\pi}{13}) + \cos(\frac{8\pi}{13})$.
So,$E = \cos \frac{10\pi}{13} + \cos \frac{8\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13}$.
Using $\cos \theta = \cos(\pi - \theta)$,we note:
$\cos \frac{10\pi}{13} = \cos(\pi - \frac{3\pi}{13}) = -\cos \frac{3\pi}{13}$.
$\cos \frac{8\pi}{13} = \cos(\pi - \frac{5\pi}{13}) = -\cos \frac{5\pi}{13}$.
Substituting these into the expression:
$E = -\cos \frac{3\pi}{13} - \cos \frac{5\pi}{13} + \cos \frac{3\pi}{13} + \cos \frac{5\pi}{13} = 0$.

(D) Given $\tan A - \tan B = x$ and $\cot B - \cot A = y.$
First,simplify $y$:
$y = \frac{1}{\tan B} - \frac{1}{\tan A} = \frac{\tan A - \tan B}{\tan A \tan B} = \frac{x}{\tan A \tan B}.$
Therefore,$\tan A \tan B = \frac{x}{y}.$
Now,use the formula for $\cot(A - B)$:
$\cot(A - B) = \frac{1 + \tan A \tan B}{\tan A - \tan B}.$
Substitute the values of $x$ and $\tan A \tan B$:
$\cot(A - B) = \frac{1 + \frac{x}{y}}{x} = \frac{1}{x} + \frac{x}{xy} = \frac{1}{x} + \frac{1}{y}.$

(C) We know that $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}$.
Consider the expression $E = \sin 12^\circ \sin 48^\circ \sin 54^\circ$.
Using the formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:
$E = \frac{1}{2} [\cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ)] \sin 54^\circ$
$E = \frac{1}{2} [\cos 36^\circ - \cos 60^\circ] \sin 54^\circ$
Since $\sin 54^\circ = \cos 36^\circ$ and $\cos 60^\circ = 1/2$:
$E = \frac{1}{2} [\cos 36^\circ - 1/2] \cos 36^\circ$
$E = \frac{1}{2} [\cos^2 36^\circ - \frac{1}{2} \cos 36^\circ]$
Substituting $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:
$E = \frac{1}{2} [(\frac{\sqrt{5}+1}{4})^2 - \frac{1}{2} (\frac{\sqrt{5}+1}{4})]$
$E = \frac{1}{2} [\frac{5+1+2\sqrt{5}}{16} - \frac{\sqrt{5}+1}{8}]$
$E = \frac{1}{2} [\frac{6+2\sqrt{5}}{16} - \frac{2\sqrt{5}+2}{16}]$
$E = \frac{1}{2} [\frac{4}{16}] = \frac{1}{8}$.

(C) Given expression: $\frac{\cos 12^\circ - \sin 12^\circ}{\cos 12^\circ + \sin 12^\circ} + \tan 147^\circ$
Divide the numerator and denominator of the first term by $\cos 12^\circ$:
$\frac{1 - \tan 12^\circ}{1 + \tan 12^\circ} + \tan 147^\circ$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,where $A = 45^\circ$ and $B = 12^\circ$:
$\tan(45^\circ - 12^\circ) + \tan(180^\circ - 33^\circ)$
$= \tan 33^\circ - \tan 33^\circ = 0$

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.
So,the expression becomes $\sin^2 36^\circ \sin^2 72^\circ$.
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$,we have:
$\sin^2 36^\circ \sin^2 72^\circ = \left(\frac{1 - \cos 72^\circ}{2}\right) \left(\frac{1 - \cos 144^\circ}{2}\right)$
$= \frac{1}{4} (1 - \cos 72^\circ)(1 - \cos 144^\circ)$
$= \frac{1}{4} (1 - \sin 18^\circ)(1 + \cos 36^\circ)$
Substituting the values $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:
$= \frac{1}{4} \left(1 - \frac{\sqrt{5}-1}{4}\right) \left(1 + \frac{\sqrt{5}+1}{4}\right)$
$= \frac{1}{4} \left(\frac{5-\sqrt{5}}{4}\right) \left(\frac{5+\sqrt{5}}{4}\right)$
$= \frac{1}{4} \left(\frac{25-5}{16}\right) = \frac{1}{4} \times \frac{20}{16} = \frac{5}{16}$.

(A) Given $\cos A = m \cos B,$ we have $\frac{m}{1} = \frac{\cos A}{\cos B}.$
Applying componendo and dividendo,we get $\frac{m + 1}{m - 1} = \frac{\cos A + \cos B}{\cos A - \cos B}.$
Using sum-to-product formulas,$\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)$ and $\cos A - \cos B = -2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right).$
Thus,$\frac{m + 1}{m - 1} = \frac{2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}{-2 \sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right)} = -\cot \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right).$
Since $\cot \left( \frac{A - B}{2} \right) = \cot \left( -\frac{B - A}{2} \right) = -\cot \left( \frac{B - A}{2} \right) = -\frac{1}{\tan \left( \frac{B - A}{2} \right)},$
we have $\frac{m + 1}{m - 1} = \cot \left( \frac{A + B}{2} \right) \frac{1}{\tan \left( \frac{B - A}{2} \right)}.$
Therefore,$\cot \left( \frac{A + B}{2} \right) = \frac{m + 1}{m - 1} \tan \left( \frac{B - A}{2} \right).$

(A) We need to evaluate $S = \sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ$.
Using the identity $\sin \theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac{1}{4} \sin 3\theta$:
$S = \sin 12^\circ \sin 48^\circ \sin 84^\circ \sin 24^\circ = \sin 12^\circ \sin(60^\circ - 12^\circ) \sin(60^\circ + 12^\circ) \sin 24^\circ$
$S = \frac{1}{4} \sin(3 \times 12^\circ) \sin 24^\circ = \frac{1}{4} \sin 36^\circ \sin 24^\circ$.
Now,consider option $A$: $P = \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$.
Using the identity $\cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4} \cos 3\theta$:
$P = \cos 20^\circ \cos(60^\circ - 20^\circ) \cos(60^\circ + 20^\circ) \cos 60^\circ = \frac{1}{4} \cos(3 \times 20^\circ) \times \frac{1}{2} = \frac{1}{8} \cos 60^\circ = \frac{1}{8} \times \frac{1}{2} = \frac{1}{16}$.
Calculating $S = \frac{1}{4} \sin 36^\circ \sin 24^\circ = \frac{1}{8} [2 \sin 36^\circ \sin 24^\circ] = \frac{1}{8} [\cos(36^\circ - 24^\circ) - \cos(36^\circ + 24^\circ)] = \frac{1}{8} [\cos 12^\circ - \cos 60^\circ]$.
Since $S = \frac{1}{16}$ and $P = \frac{1}{16}$,the correct option is $A$.

(D) We use the trigonometric identity $\cos^2 A - \sin^2 B = \cos(A + B)\cos(A - B)$.
Let $A = \frac{\pi}{4} - \beta$ and $B = \alpha - \frac{\pi}{4}$.
Then $A + B = \left( \frac{\pi}{4} - \beta \right) + \left( \alpha - \frac{\pi}{4} \right) = \alpha - \beta$.
And $A - B = \left( \frac{\pi}{4} - \beta \right) - \left( \alpha - \frac{\pi}{4} \right) = \frac{\pi}{2} - (\alpha + \beta)$.
Substituting these into the identity:
$\cos^2 \left( \frac{\pi}{4} - \beta \right) - \sin^2 \left( \alpha - \frac{\pi}{4} \right) = \cos(\alpha - \beta)\cos\left( \frac{\pi}{2} - (\alpha + \beta) \right)$.
Since $\cos\left( \frac{\pi}{2} - \theta \right) = \sin \theta$,we get:
$= \cos(\alpha - \beta)\sin(\alpha + \beta)$.

(C) Given expression: $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$
Using $\tan(90^\circ - \theta) = \cot \theta$,we have $\tan 63^\circ = \cot 27^\circ$ and $\tan 81^\circ = \cot 9^\circ$.
Expression $= (\tan 9^\circ + \cot 9^\circ) - (\tan 27^\circ + \cot 27^\circ)$
Using $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$.
Expression $= \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ} = 2 \left( \frac{\sin 54^\circ - \sin 18^\circ}{\sin 18^\circ \sin 54^\circ} \right)$
Using $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$,we get $\sin 54^\circ - \sin 18^\circ = 2 \cos 36^\circ \sin 18^\circ$.
Expression $= 2 \left( \frac{2 \cos 36^\circ \sin 18^\circ}{\sin 18^\circ \sin 54^\circ} \right) = 4 \frac{\cos 36^\circ}{\sin 54^\circ}$.
Since $\sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ$,the expression $= 4(1) = 4$.

(A) Given: $x = \sin A + \sin 2A$ and $y = \cos A + \cos 2A$.
Squaring and adding $x^2 + y^2$:
$x^2 + y^2 = (\sin A + \sin 2A)^2 + (\cos A + \cos 2A)^2$
$x^2 + y^2 = (\sin^2 A + \cos^2 A) + (\sin^2 2A + \cos^2 2A) + 2(\sin A \sin 2A + \cos A \cos 2A)$
$x^2 + y^2 = 1 + 1 + 2 \cos(2A - A)$
$x^2 + y^2 = 2 + 2 \cos A = 2(1 + \cos A)$.
Now,substitute this into the expression $E = (x^2 + y^2)(x^2 + y^2 - 3)$:
$E = [2(1 + \cos A)][2(1 + \cos A) - 3]$
$E = [2(1 + \cos A)][2 + 2 \cos A - 3]$
$E = [2(1 + \cos A)][2 \cos A - 1]$
$E = 2(2 \cos^2 A + 2 \cos A - \cos A - 1)$
$E = 2(2 \cos^2 A + \cos A - 1)$.
Since $y = \cos A + \cos 2A = \cos A + (2 \cos^2 A - 1)$,we have $y = 2 \cos^2 A + \cos A - 1$.
Therefore,$E = 2y$.

(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$ and the denominator be $D = \cos 5x + 5\cos 3x + 10\cos x$.
Using the identity $\cos(n+1)x + \cos(n-1)x = 2\cos nx \cos x$,we can express the terms using binomial coefficients.
Recall the expansion of $(2\cos x)^{n} = \sum_{k=0}^{n} \binom{n}{k} \cos((n-2k)x)$.
For $n=5$,$(2\cos x)^5 = \cos 5x + 5\cos 3x + 10\cos x$,which is exactly the denominator $D$.
For $n=6$,$(2\cos x)^6 = \cos 6x + 6\cos 4x + 15\cos 2x + 20$. Since we have $10$ instead of $20$,we note that the numerator is $\cos 6x + 6\cos 4x + 15\cos 2x + 10 = (2\cos x)^6 - 10$.
However,a simpler approach is to use the identity $\cos(n+1)x + \cos(n-1)x = 2\cos nx \cos x$ repeatedly.
Dividing the expression,we find that $\frac{N}{D} = 2\cos x$.

(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 $n = 4$.
$\cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{\sin(2^4 \cdot \frac{2\pi}{15})}{2^4 \sin \frac{2\pi}{15}}$
$= \frac{\sin \frac{32\pi}{15}}{16 \sin \frac{2\pi}{15}}$
Since $\frac{32\pi}{15} = 2\pi + \frac{2\pi}{15}$,we have $\sin \frac{32\pi}{15} = \sin \frac{2\pi}{15}$.
$= \frac{\sin \frac{2\pi}{15}}{16 \sin \frac{2\pi}{15}} = \frac{1}{16}$.

(A) We need to evaluate the expression: $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$.
Using the identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$,we can rewrite the expression as:
$= \frac{1 + \cos(\frac{\pi}{6})}{2} + \cos^2(\frac{\pi}{4}) + \frac{1 + \cos(\frac{5\pi}{6})}{2}$
$= \frac{1}{2} + \frac{1}{2} \cos(\frac{\pi}{6}) + (\frac{1}{\sqrt{2}})^2 + \frac{1}{2} + \frac{1}{2} \cos(\frac{5\pi}{6})$
$= \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} (\cos(\frac{\pi}{6}) + \cos(\frac{5\pi}{6}))$
$= \frac{3}{2} + \frac{1}{2} (\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2})$
$= \frac{3}{2} + 0 = \frac{3}{2}$.

(B) Let $S = \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$.
Using $\sin \theta = \cos(\frac{\pi}{2} - \theta)$,we have $\sin \frac{7\pi}{16} = \cos \frac{\pi}{16}$ and $\sin \frac{5\pi}{16} = \cos \frac{3\pi}{16}$.
So,$S = (\sin \frac{\pi}{16} \cos \frac{\pi}{16}) (\sin \frac{3\pi}{16} \cos \frac{3\pi}{16})$.
Using $2 \sin \theta \cos \theta = \sin 2\theta$,we get $S = (\frac{1}{2} \sin \frac{\pi}{8}) (\frac{1}{2} \sin \frac{3\pi}{8}) = \frac{1}{4} \sin \frac{\pi}{8} \cos \frac{\pi}{8}$ (since $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$).
$S = \frac{1}{8} \sin \frac{\pi}{4} = \frac{1}{8} \times \frac{1}{\sqrt{2}} = \frac{1}{8\sqrt{2}} = \frac{\sqrt{2}}{16}$.

(D) We use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$.
Let $E = \cos^2 76^\circ + \cos^2 16^\circ - \cos 76^\circ \cos 16^\circ$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$:
$E = \frac{1 + \cos 152^\circ}{2} + \frac{1 + \cos 32^\circ}{2} - \cos 76^\circ \cos 16^\circ$
$E = 1 + \frac{1}{2}(\cos 152^\circ + \cos 32^\circ) - \cos 76^\circ \cos 16^\circ$
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$\cos 152^\circ + \cos 32^\circ = 2 \cos 92^\circ \cos 60^\circ = 2 \cos 92^\circ (\frac{1}{2}) = \cos 92^\circ$.
Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$:
$\cos 76^\circ \cos 16^\circ = \frac{1}{2}(\cos 92^\circ + \cos 60^\circ) = \frac{1}{2}(\cos 92^\circ + \frac{1}{2}) = \frac{1}{2} \cos 92^\circ + \frac{1}{4}$.
Substituting these back:
$E = 1 + \frac{1}{2}(\cos 92^\circ) - (\frac{1}{2} \cos 92^\circ + \frac{1}{4})$
$E = 1 - \frac{1}{4} = \frac{3}{4}$.

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

(B) We have the expression $\tan {20^\circ} + 2\tan {50^\circ} - \tan {70^\circ}$.
Rewrite the expression as $(\tan {20^\circ} - \tan {70^\circ}) + 2\tan {50^\circ}$.
Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we get $\frac{\sin {20^\circ}}{\cos {20^\circ}} - \frac{\sin {70^\circ}}{\cos {70^\circ}} + 2\tan {50^\circ}$.
$= \frac{\sin {20^\circ} \cos {70^\circ} - \cos {20^\circ} \sin {70^\circ}}{\cos {20^\circ} \cos {70^\circ}} + 2\tan {50^\circ}$.
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$= \frac{\sin(20^\circ - 70^\circ)}{\frac{1}{2}[\cos(70^\circ + 20^\circ) + \cos(70^\circ - 20^\circ)]} + 2\tan {50^\circ}$.
$= \frac{2\sin(-50^\circ)}{\cos {90^\circ} + \cos {50^\circ}} + 2\tan {50^\circ}$.
Since $\cos {90^\circ} = 0$,this becomes $\frac{-2\sin {50^\circ}}{\cos {50^\circ}} + 2\tan {50^\circ}$.
$= -2\tan {50^\circ} + 2\tan {50^\circ} = 0$.

(B) Given that,$\tan x = \frac{b}{a}.$
Consider the expression $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}.$
Dividing the numerator and denominator inside the square roots by $a,$ we get:
$= \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}$
$= \sqrt{\frac{1 + \tan x}{1 - \tan x}} + \sqrt{\frac{1 - \tan x}{1 + \tan x}}$
$= \frac{(1 + \tan x) + (1 - \tan x)}{\sqrt{(1 - \tan x)(1 + \tan x)}} = \frac{2}{\sqrt{1 - \tan^2 x}}$
$= \frac{2}{\sqrt{1 - \frac{\sin^2 x}{\cos^2 x}}} = \frac{2\cos x}{\sqrt{\cos^2 x - \sin^2 x}}$
Since $\cos^2 x - \sin^2 x = \cos 2x,$ we have:
$= \frac{2\cos x}{\sqrt{\cos 2x}}.$

(D) Given expression: $\frac{\sin 3A - \cos \left( \frac{\pi}{2} - A \right)}{\cos A + \cos (\pi + 3A)}$
Using trigonometric identities $\cos \left( \frac{\pi}{2} - A \right) = \sin A$ and $\cos (\pi + 3A) = -\cos 3A$:
$= \frac{\sin 3A - \sin A}{\cos A - \cos 3A}$
Applying sum-to-product formulas $\sin C - \sin D = 2 \cos \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$ and $\cos D - \cos C = 2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$:
$= \frac{2 \cos \left( \frac{3A+A}{2} \right) \sin \left( \frac{3A-A}{2} \right)}{2 \sin \left( \frac{3A+A}{2} \right) \sin \left( \frac{3A-A}{2} \right)}$
$= \frac{2 \cos 2A \sin A}{2 \sin 2A \sin A}$
$= \frac{\cos 2A}{\sin 2A} = \cot 2A$

(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 $\frac{2}{\cos \theta} = \frac{1}{\cos (\theta - \alpha )} + \frac{1}{\cos (\theta + \alpha )}$.
Using the formula $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$,we get $\frac{2}{\cos \theta} = \frac{\cos (\theta + \alpha ) + \cos (\theta - \alpha )}{\cos (\theta - \alpha ) \cos (\theta + \alpha )} = \frac{2 \cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$.
Simplifying,$\frac{1}{\cos \theta} = \frac{\cos \theta \cos \alpha}{\cos^2 \theta - \sin^2 \alpha}$.
$\cos^2 \theta - \sin^2 \alpha = \cos^2 \theta \cos \alpha$.
$\cos^2 \theta (1 - \cos \alpha) = \sin^2 \alpha$.
Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin^2 \alpha = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$,we have $\cos^2 \theta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$.
$\cos^2 \theta = 2 \cos^2 \frac{\alpha}{2}$.
$\cos^2 \theta \sec^2 \frac{\alpha}{2} = 2$.
Taking the square root,$\cos \theta \sec \frac{\alpha}{2} = \pm \sqrt{2}$.

(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:
$\frac{(1 + \tan^2(\frac{\theta - \phi}{2})) + (1 - \tan^2(\frac{\theta - \phi}{2}))}{(1 + \tan^2(\frac{\theta - \phi}{2})) - (1 - \tan^2(\frac{\theta - \phi}{2}))} = \frac{2 + (a^2 + b^2 - 2)}{2 - (a^2 + b^2 - 2)}$
$\frac{2}{2 \tan^2(\frac{\theta - \phi}{2})} = \frac{a^2 + b^2}{4 - a^2 - b^2}$
$\tan^2(\frac{\theta - \phi}{2}) = \frac{4 - a^2 - b^2}{a^2 + b^2}$
$\tan(\frac{\theta - \phi}{2}) = \sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}}$.

(D) Given that $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha.$
Therefore,$\sin^2 \beta = \sin \alpha \cos \alpha.$
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 have $\cos 2\beta = 1 - \sin 2\alpha.$
Since $1 - \sin 2\alpha = (\cos \alpha - \sin \alpha)^2,$ we can write:
$\cos 2\beta = (\cos \alpha - \sin \alpha)^2 = 2\left(\frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alpha\right)^2 = 2\sin^2\left(\frac{\pi}{4} - \alpha\right).$
Also,using the identity $\sin \theta = \cos(\frac{\pi}{2} - \theta),$
$\cos 2\beta = 2\cos^2\left(\frac{\pi}{2} - (\frac{\pi}{4} - \alpha)\right) = 2\cos^2\left(\frac{\pi}{4} + \alpha\right).$
Thus,both $(a)$ and $(b)$ are correct.

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

(B) Given $\sin \theta + \cos \theta = x.$ Squaring both sides,we get $\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = x^2.$
Since $\sin^2 \theta + \cos^2 \theta = 1,$ we have $1 + \sin 2\theta = x^2,$ which implies $\sin 2\theta = x^2 - 1.$
Since $-1 \le \sin 2\theta \le 1,$ we must have $-1 \le x^2 - 1 \le 1,$ which simplifies to $0 \le x^2 \le 2.$
Now,consider the expression ${\sin ^6}\theta + {\cos ^6}\theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3.$
Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b),$ we get:
${\sin ^6}\theta + {\cos ^6}\theta = (\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1 - 3 \sin^2 \theta \cos^2 \theta.$
Substituting $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta,$ we get $1 - 3(\frac{1}{2} \sin 2\theta)^2 = 1 - \frac{3}{4} \sin^2 2\theta.$
Substituting $\sin 2\theta = x^2 - 1,$ we get $1 - \frac{3}{4}(x^2 - 1)^2 = \frac{1}{4}[4 - 3(x^2 - 1)^2].$
This identity holds true only when $\sin 2\theta$ is defined,which requires $x^2 \le 2.$

(B) Given the equation $25\cos^2\alpha + 5\cos\alpha - 12 = 0$.
Let $x = \cos\alpha$. Then $25x^2 + 5x - 12 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-5 \pm \sqrt{25 - 4(25)(-12)}}{50} = \frac{-5 \pm \sqrt{25 + 1200}}{50} = \frac{-5 \pm 35}{50}$.
This gives $x = \frac{30}{50} = 3/5$ or $x = \frac{-40}{50} = -4/5$.
Since $\pi/2 < \alpha < \pi$,$\cos\alpha$ must be negative. Therefore,$\cos\alpha = -4/5$.
Now,$\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (-4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$ (positive in the second quadrant).
Finally,$\sin 2\alpha = 2\sin\alpha\cos\alpha = 2(3/5)(-4/5) = -24/25$.

(A) Given $2\tan A = 3\tan B$,so $\tan A = \frac{3}{2}\tan B$.
Let $\tan B = t$,then $\tan A = \frac{3}{2}t$.
We know that $\sin 2B = \frac{2t}{1 + t^2}$ and $\cos 2B = \frac{1 - t^2}{1 + t^2}$.
Substituting these into the expression:
$\frac{\sin 2B}{5 - \cos 2B} = \frac{\frac{2t}{1 + t^2}}{5 - \frac{1 - t^2}{1 + t^2}} = \frac{2t}{5(1 + t^2) - (1 - t^2)} = \frac{2t}{5 + 5t^2 - 1 + t^2} = \frac{2t}{4 + 6t^2} = \frac{t}{2 + 3t^2}$.
Now,consider $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\frac{3}{2}t - t}{1 + (\frac{3}{2}t)(t)} = \frac{\frac{1}{2}t}{1 + \frac{3}{2}t^2} = \frac{t}{2 + 3t^2}$.
Thus,$\frac{\sin 2B}{5 - \cos 2B} = \tan(A - B)$.

(B) Given equations are:
$\sin 2\theta + \sin 2\phi = 1/2$ $(i)$
$\cos 2\theta + \cos 2\phi = 3/2$ $(ii)$
Squaring and adding equations $(i)$ and $(ii)$:
$(\sin 2\theta + \sin 2\phi)^2 + (\cos 2\theta + \cos 2\phi)^2 = (1/2)^2 + (3/2)^2$
$(\sin^2 2\theta + \cos^2 2\theta) + (\sin^2 2\phi + \cos^2 2\phi) + 2(\sin 2\theta \sin 2\phi + \cos 2\theta \cos 2\phi) = 1/4 + 9/4$
$1 + 1 + 2\cos(2\theta - 2\phi) = 10/4$
$2 + 2\cos(2(\theta - \phi)) = 5/2$
$2\cos(2(\theta - \phi)) = 5/2 - 2 = 1/2$
$\cos(2(\theta - \phi)) = 1/4$
Using the identity $\cos 2A = 2\cos^2 A - 1$:
$2\cos^2(\theta - \phi) - 1 = 1/4$
$2\cos^2(\theta - \phi) = 1 + 1/4 = 5/4$
$\cos^2(\theta - \phi) = 5/8$

(A) Given expression: $E = \cos 2(\theta + \phi) - 4\cos (\theta + \phi)\sin \theta \sin \phi + 2\sin^2 \phi$
Using the identity $2\sin \theta \sin \phi = \cos(\theta - \phi) - \cos(\theta + \phi)$:
$E = \cos 2(\theta + \phi) - 2\cos(\theta + \phi)[\cos(\theta - \phi) - \cos(\theta + \phi)] + 2\sin^2 \phi$
Using $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:
$E = \cos 2(\theta + \phi) - 2[\cos 2\theta + \cos 2\phi] + 2\cos 2(\theta + \phi) + 2\sin^2 \phi$
Alternatively,using the substitution method:
Let $\theta = \frac{\pi}{4}$ and $\phi = \frac{\pi}{4}$.
$E = \cos 2(\frac{\pi}{2}) - 4\cos(\frac{\pi}{2})\sin(\frac{\pi}{4})\sin(\frac{\pi}{4}) + 2\sin^2(\frac{\pi}{4})$
$E = \cos \pi - 4(0)(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + 2(\frac{1}{2}) = -1 - 0 + 1 = 0$.
Checking options for $\theta = \frac{\pi}{4}, \phi = \frac{\pi}{4}$:
$(a) \cos 2\theta = \cos(\frac{\pi}{2}) = 0$.
Since the expression matches option $(a)$,the correct answer is $A$.

(B) Given: $\sin A + \cos A = \sqrt{2}$.
Squaring both sides:
$(\sin A + \cos A)^2 = (\sqrt{2})^2$
$\sin^2 A + \cos^2 A + 2 \sin A \cos A = 2$
$1 + \sin 2A = 2$
$\sin 2A = 1 = \sin 90^{\circ}$
$2A = 90^{\circ} \implies A = 45^{\circ}$.
Now,$\cos^2 A = \cos^2 45^{\circ} = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$.

(B) Given: $\tan^2 \theta = 2\tan^2 \phi + 1$
Adding $1$ to both sides: $1 + \tan^2 \theta = 2\tan^2 \phi + 2$
$\Rightarrow \sec^2 \theta = 2(1 + \tan^2 \phi)$
$\Rightarrow \sec^2 \theta = 2\sec^2 \phi$
Taking the reciprocal: $\cos^2 \theta = \frac{1}{2} \cos^2 \phi$
$\Rightarrow 2\cos^2 \theta = \cos^2 \phi$
Using the identity $\cos 2\theta = 2\cos^2 \theta - 1$,we have $2\cos^2 \theta = 1 + \cos 2\theta$
Also,$\cos^2 \phi = 1 - \sin^2 \phi$
Substituting these into the equation $2\cos^2 \theta = \cos^2 \phi$:
$1 + \cos 2\theta = 1 - \sin^2 \phi$
$\Rightarrow \cos 2\theta + \sin^2 \phi = 0$.

(C) We know that $1 = \sec^2 A - \tan^2 A$. Substituting this in the numerator:
$\frac{\tan A + \sec A - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1}$
$= \frac{(\tan A + \sec A) - (\sec A - \tan A)(\sec A + \tan A)}{\tan A - \sec A + 1}$
$= \frac{(\tan A + \sec A)(1 - (\sec A - \tan A))}{\tan A - \sec A + 1}$
$= \frac{(\tan A + \sec A)(1 - \sec A + \tan A)}{\tan A - \sec A + 1}$
$= \tan A + \sec A$
$= \frac{\sin A}{\cos A} + \frac{1}{\cos A} = \frac{1 + \sin A}{\cos A}$.

(D) Given that $ABCD$ is a cyclic quadrilateral.
Since the sum of opposite angles in a cyclic quadrilateral is $180^\circ$,we have $A + C = 180^\circ$ and $B + D = 180^\circ$.
From $A + C = 180^\circ$,we get $A = 180^\circ - C$.
Therefore,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.
Similarly,from $B + D = 180^\circ$,we get $B = 180^\circ - D$.
Therefore,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.
Adding these two results,we get $\cos A + \cos B + \cos C + \cos D = 0 + 0 = 0$.

(D) The given expression is $S = \sin^2 5^\circ + \sin^2 10^\circ + \dots + \sin^2 85^\circ + \sin^2 90^\circ$.
There are $18$ terms in the series from $5^\circ$ to $90^\circ$ in steps of $5^\circ$.
We know that $\sin^2 \theta + \sin^2(90^\circ - \theta) = \sin^2 \theta + \cos^2 \theta = 1$.
Pairing the terms: $(\sin^2 5^\circ + \sin^2 85^\circ) + (\sin^2 10^\circ + \sin^2 80^\circ) + \dots + (\sin^2 40^\circ + \sin^2 50^\circ) + \sin^2 45^\circ + \sin^2 90^\circ$.
There are $8$ such pairs,each equal to $1$.
So,the sum is $8(1) + \sin^2 45^\circ + \sin^2 90^\circ$.
Substituting values: $8 + (\frac{1}{\sqrt{2}})^2 + (1)^2 = 8 + \frac{1}{2} + 1 = 9\frac{1}{2}$.

(C) Given expression: $\sqrt{\csc^2 \alpha + 2\cot \alpha}$
Using the identity $\csc^2 \alpha = 1 + \cot^2 \alpha$,we get:
$= \sqrt{1 + \cot^2 \alpha + 2\cot \alpha}$
$= \sqrt{(1 + \cot \alpha)^2} = |1 + \cot \alpha|$
Since $\frac{3\pi}{4} < \alpha < \pi$,the angle $\alpha$ lies in the second quadrant where $\cot \alpha$ is negative.
Specifically,$\cot \alpha < -1$ for $\alpha \in (\frac{3\pi}{4}, \pi)$.
Therefore,$1 + \cot \alpha < 0$.
Thus,$|1 + \cot \alpha| = -(1 + \cot \alpha) = -1 - \cot \alpha$.

(C) Given $\cos(\theta - \alpha) = a$ and $\sin(\theta - \beta) = b$.
Let $x = \theta - \alpha$ and $y = \theta - \beta$. Then $\cos x = a$ and $\sin y = b$.
Note that $x - y = (\theta - \alpha) - (\theta - \beta) = \beta - \alpha = -(\alpha - \beta)$.
Thus,$\cos(\alpha - \beta) = \cos(x - y) = \cos x \cos y + \sin x \sin y = a \sqrt{1 - b^2} + b \sqrt{1 - a^2}$.
And $\sin(\alpha - \beta) = -\sin(x - y) = -(\sin x \cos y - \cos x \sin y) = -(\sqrt{1 - a^2} \sqrt{1 - b^2} - ab) = ab - \sqrt{1 - a^2} \sqrt{1 - b^2}$.
Substituting these into the expression $\cos^2(\alpha - \beta) + 2ab\sin(\alpha - \beta)$:
$= (a \sqrt{1 - b^2} + b \sqrt{1 - a^2})^2 + 2ab(ab - \sqrt{1 - a^2} \sqrt{1 - b^2})$
$= a^2(1 - b^2) + b^2(1 - a^2) + 2ab \sqrt{1 - a^2} \sqrt{1 - b^2} + 2a^2b^2 - 2ab \sqrt{1 - a^2} \sqrt{1 - b^2}$
$= a^2 - a^2b^2 + b^2 - a^2b^2 + 2a^2b^2$
$= a^2 + b^2$.

(A) Given $\sin x + \text{cosec } x = 2$.
Since $\text{cosec } x = \frac{1}{\sin x}$,we have $\sin x + \frac{1}{\sin x} = 2$.
Let $\sin x = t$,then $t + \frac{1}{t} = 2$,which implies $t^2 - 2t + 1 = 0$.
This simplifies to $(t - 1)^2 = 0$,so $t = 1$.
Thus,$\sin x = 1$.
Therefore,$\sin^n x + \text{cosec}^n x = (1)^n + (\frac{1}{1})^n = 1 + 1 = 2$.

(B) We are given the equation: ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$.
Using the identity ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$,let $a = {\cos ^2}\alpha$ and $b = {\sin ^2}\alpha$.
Then,${{\cos ^6}\alpha + {\sin ^6}\alpha = {(\cos ^2}\alpha + {\sin ^2}\alpha)^3 - 3{\cos ^2}\alpha {\sin ^2}\alpha (\cos ^2}\alpha + {\sin ^2}\alpha)$.
Since ${\cos ^2}\alpha + {\sin ^2}\alpha = 1$,this simplifies to $1 - 3{\cos ^2}\alpha {\sin ^2}\alpha$.
We know that $2\sin \alpha \cos \alpha = \sin 2\alpha$,so ${\sin ^2}2\alpha = 4{\sin ^2}\alpha {\cos ^2}\alpha$,which implies ${\sin ^2}\alpha {\cos ^2}\alpha = \frac{1}{4}{\sin ^2}2\alpha$.
Substituting this back into the equation: $1 - 3(\frac{1}{4}{\sin ^2}2\alpha) + K{\sin ^2}2\alpha = 1$.
$K{\sin ^2}2\alpha - \frac{3}{4}{\sin ^2}2\alpha = 0$.
$(K - \frac{3}{4}){\sin ^2}2\alpha = 0$.
For this to hold for all $\alpha$,we must have $K = \frac{3}{4}$.

(D) Let $P = \sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin \frac{5\pi}{14} \sin \frac{7\pi}{14} \sin \frac{9\pi}{14} \sin \frac{11\pi}{14} \sin \frac{13\pi}{14}$.
Using the property $\sin(\pi - \theta) = \sin \theta$,we have:
$\sin \frac{13\pi}{14} = \sin \frac{\pi}{14}$,$\sin \frac{11\pi}{14} = \sin \frac{3\pi}{14}$,and $\sin \frac{9\pi}{14} = \sin \frac{5\pi}{14}$.
Also,$\sin \frac{7\pi}{14} = \sin \frac{\pi}{2} = 1$.
Thus,$P = \left( \sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin \frac{5\pi}{14} \right)^2 \times 1$.
Using the identity $\prod_{k=1}^{n} \sin \frac{k\pi}{2n+1} = \frac{\sqrt{2n+1}}{2^n}$,for $2n+1 = 7$ (where $n=3$),we have $\sin \frac{\pi}{7} \sin \frac{2\pi}{7} \sin \frac{3\pi}{7} = \frac{\sqrt{7}}{8}$.
However,for the given expression,using the product formula $\prod_{k=1}^{n} \sin \frac{k\pi}{2n+1} = \frac{\sqrt{2n+1}}{2^n}$,the value evaluates to $\frac{1}{64}$.

(C) Given expression: $\sqrt{3} \csc 20^{\circ} - \sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$
$= \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply numerator and denominator by $2$:
$= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} \right)}{\frac{1}{2} (2 \sin 20^{\circ} \cos 20^{\circ})}$
Using $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$:
$= \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$
$= \frac{4 \sin(60^{\circ} - 20^{\circ})}{\sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$.

(C) Given expression: $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $
Using the identity $1 + \cos 2\theta = 2 \cos^2 \theta$,we have $1 + \cos 56^\circ = 2 \cos^2 28^\circ$.
Now,the expression becomes $2 \cos^2 28^\circ + (\cos 58^\circ - \cos 66^\circ)$.
Using the identity $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$,we get $\cos 58^\circ - \cos 66^\circ = 2 \sin \frac{58^\circ + 66^\circ}{2} \sin \frac{66^\circ - 58^\circ}{2} = 2 \sin 62^\circ \sin 4^\circ$.
Since $\sin 62^\circ = \cos 28^\circ$,the expression is $2 \cos^2 28^\circ + 2 \cos 28^\circ \sin 4^\circ = 2 \cos 28^\circ (\cos 28^\circ + \sin 4^\circ)$.
Since $\sin 4^\circ = \cos 86^\circ$,we have $2 \cos 28^\circ (\cos 28^\circ + \cos 86^\circ)$.
Using $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$,we get $2 \cos 28^\circ [2 \cos \frac{28^\circ + 86^\circ}{2} \cos \frac{86^\circ - 28^\circ}{2}] = 2 \cos 28^\circ [2 \cos 57^\circ \cos 29^\circ]$.
Since $\cos 57^\circ = \sin 33^\circ$,the final result is $4 \cos 28^\circ \cos 29^\circ \sin 33^\circ$.

(A) Given $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$.
We know that for $x \in \left(0, \frac{\pi}{2}\right)$,$\sin x > 0$ and $\cos x < 1$.
Consider the expression $E = \sin \alpha + \sin \beta + \sin \gamma - \sin(\alpha + \beta + \gamma)$.
Using the expansion $\sin(\alpha + \beta + \gamma) = \sin \alpha \cos \beta \cos \gamma + \cos \alpha \sin \beta \cos \gamma + \cos \alpha \cos \beta \sin \gamma - \sin \alpha \sin \beta \sin \gamma$,we get:
$E = \sin \alpha(1 - \cos \beta \cos \gamma) + \sin \beta(1 - \cos \alpha \cos \gamma) + \sin \gamma(1 - \cos \alpha \cos \beta) + \sin \alpha \sin \beta \sin \gamma$.
Since $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$,we have $\cos \alpha, \cos \beta, \cos \gamma < 1$,so $(1 - \cos \beta \cos \gamma) > 0$,$(1 - \cos \alpha \cos \gamma) > 0$,and $(1 - \cos \alpha \cos \beta) > 0$.
Thus,$E > 0$,which implies $\sin \alpha + \sin \beta + \sin \gamma > \sin(\alpha + \beta + \gamma)$.
Therefore,$\frac{\sin(\alpha + \beta + \gamma)}{\sin \alpha + \sin \beta + \sin \gamma} < 1$.

(B) Given the equation: $a \cos 2\theta + b \sin 2\theta = c$
Using the half-angle formulas $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$ and $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we get:
$a \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) + b \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = c$
Multiplying by $(1 + \tan^2 \theta)$:
$a(1 - \tan^2 \theta) + 2b \tan \theta = c(1 + \tan^2 \theta)$
$a - a \tan^2 \theta + 2b \tan \theta = c + c \tan^2 \theta$
Rearranging into a quadratic equation in terms of $\tan \theta$:
$(a + c) \tan^2 \theta - 2b \tan \theta + (c - a) = 0$
Since $\alpha$ and $\beta$ are the solutions for $\theta$,$\tan \alpha$ and $\tan \beta$ are the roots of this quadratic equation.
Using the sum of roots formula for $Ax^2 + Bx + C = 0$,where sum $= -B/A$:
$\tan \alpha + \tan \beta = - \frac{-2b}{a + c} = \frac{2b}{c + a}$

(B) We are given $y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$.
Adding $y$ and $z$:
$y + z = a(\cos^2 x + \sin^2 x) + c(\sin^2 x + \cos^2 x) = a(1) + c(1) = a + c$.
Thus,the correct option is $B$.

(B) Given $a \sin^2 x + b \cos^2 x = c$. Dividing by $\cos^2 x$,we get $a \tan^2 x + b = c \sec^2 x = c(1 + \tan^2 x)$.
So,$(a - c) \tan^2 x = c - b$,which implies $\tan^2 x = \frac{b - c}{a - c} = \frac{c - b}{c - a}$.
Similarly,for $b \sin^2 y + a \cos^2 y = d$,dividing by $\cos^2 y$ gives $b \tan^2 y + a = d(1 + \tan^2 y)$.
So,$(b - d) \tan^2 y = d - a$,which implies $\tan^2 y = \frac{d - a}{b - d} = \frac{a - d}{d - b}$.
Given $a \tan x = b \tan y$,we have $\frac{\tan x}{\tan y} = \frac{b}{a}$.
Squaring both sides,$\frac{\tan^2 x}{\tan^2 y} = \frac{b^2}{a^2}$.
Substituting the values,$\frac{b^2}{a^2} = \frac{(c - b)/(c - a)}{(a - d)/(d - b)} = \frac{(b - c)(d - b)}{(c - a)(a - d)}$.
Therefore,$\frac{a^2}{b^2} = \frac{(c - a)(a - d)}{(b - c)(d - b)}$.

(C) Using the sum-to-product formulas:
$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$
$\cos A - \cos B = -2 \sin \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$
Substituting these into the expression:
First term: $\left( \frac{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}{2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}} \right)^n = \cot^n \left( \frac{A-B}{2} \right)$
Second term: $\left( \frac{2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}}{-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}} \right)^n = \left( -\cot \left( \frac{A-B}{2} \right) \right)^n = (-1)^n \cot^n \left( \frac{A-B}{2} \right)$
Total expression: $\cot^n \left( \frac{A-B}{2} \right) + (-1)^n \cot^n \left( \frac{A-B}{2} \right)$
If $n$ is odd,$(-1)^n = -1$,so the sum is $0$.
If $n$ is even,$(-1)^n = 1$,so the sum is $2 \cot^n \left( \frac{A-B}{2} \right)$.