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

(B) Given: $\cos x + \cos y = \frac{2}{3}$ $(i)$
$\sin x - \sin y = \frac{3}{4}$ $(ii)$
Using sum-to-product formulas:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = \frac{2}{3}$ $(iii)$
$2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) = \frac{3}{4}$ $(iv)$
Dividing $(iv)$ by $(iii)$:
$\tan \left(\frac{x-y}{2}\right) = \frac{3/4}{2/3} = \frac{9}{8}$
Let $\theta = \frac{x-y}{2}$,so $\tan \theta = \frac{9}{8}$.
Then $\sin(x-y) = \sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{2(9/8)}{1 + (81/64)} = \frac{9/4}{145/64} = \frac{9}{4} \times \frac{64}{145} = \frac{144}{145}$.
$\cos(x-y) = \cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - (81/64)}{1 + (81/64)} = \frac{-17/64}{145/64} = -\frac{17}{145}$.
Therefore,$\sin(x-y) + \cos(x-y) = \frac{144}{145} - \frac{17}{145} = \frac{127}{145}$.

(A) We know that $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$.
Given $\sinh x = \frac{12}{5}$,we have $\frac{e^x - e^{-x}}{2} = \frac{12}{5}$,which implies $e^x - e^{-x} = \frac{24}{5}$.
Let $e^x = t$. Then $t - \frac{1}{t} = \frac{24}{5}$.
Multiplying by $5t$,we get $5t^2 - 24t - 5 = 0$.
Factoring the quadratic equation: $(5t + 1)(t - 5) = 0$.
Since $e^x = t$ must be positive,$t = 5$,so $e^x = 5$.
We need to find $\sinh 3x + \cosh 3x$.
Using the definitions,$\sinh 3x + \cosh 3x = \frac{e^{3x} - e^{-3x}}{2} + \frac{e^{3x} + e^{-3x}}{2} = e^{3x}$.
Since $e^x = 5$,$e^{3x} = (e^x)^3 = 5^3 = 125$.

(A) Given $\cosh x = \frac{4}{3}$.
We know the identity $\cosh 2x = 2 \cosh^2 x - 1$.
$\cosh 2x = 2 \left(\frac{4}{3}\right)^2 - 1 = 2 \left(\frac{16}{9}\right) - 1 = \frac{32}{9} - 1 = \frac{23}{9}$.
We know the identity $\cosh 3x = 4 \cosh^3 x - 3 \cosh x$.
$\cosh 3x = 4 \left(\frac{4}{3}\right)^3 - 3 \left(\frac{4}{3}\right) = 4 \left(\frac{64}{27}\right) - 4 = \frac{256}{27} - 4 = \frac{256 - 108}{27} = \frac{148}{27}$.
Now,substitute these values into the expression $3 \cosh x + 9 \cosh 2x + 27 \cosh 3x$:
$= 3 \left(\frac{4}{3}\right) + 9 \left(\frac{23}{9}\right) + 27 \left(\frac{148}{27}\right)$
$= 4 + 23 + 148 = 175$.

(C) Given $A+B+C+D=2 \pi$.
We have $\cos A - \cos B + \cos C - \cos D = (\cos A - \cos B) + (\cos C - \cos D)$.
Using the formula $\cos X - \cos Y = -2 \sin \frac{X+Y}{2} \sin \frac{X-Y}{2}$,we get:
$= -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} - 2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$.
Since $A+B+C+D=2 \pi$,we have $\frac{C+D}{2} = \pi - \frac{A+B}{2}$,so $\sin \frac{C+D}{2} = \sin \frac{A+B}{2}$.
Also,$\frac{C-D}{2} = \frac{(2 \pi - A - B) - 2D}{2} = \pi - \frac{A+B+2D}{2}$.
Using the identity,the expression simplifies to:
$-4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$.

(B) Given: $\sin x \cdot \cosh y = \cos \theta$ $(i)$
$\cos x \cdot \sinh y = \sin \theta$ $(ii)$
Squaring and adding $(i)$ and $(ii)$:
$(\sin x \cdot \cosh y)^2 + (\cos x \cdot \sinh y)^2 = \cos^2 \theta + \sin^2 \theta$
$\sin^2 x \cosh^2 y + \cos^2 x \sinh^2 y = 1$
Using $\cosh^2 y = 1 + \sinh^2 y$ and $\cos^2 x = 1 - \sin^2 x$:
$\sin^2 x (1 + \sinh^2 y) + (1 - \sin^2 x) \sinh^2 y = 1$
$\sin^2 x + \sin^2 x \sinh^2 y + \sinh^2 y - \sin^2 x \sinh^2 y = 1$
$\sin^2 x + \sinh^2 y = 1$
Since $\cosh^2 y = 1 + \sinh^2 y$,we have $\sinh^2 y = \cosh^2 y - 1$.
Substituting this into the equation:
$\sin^2 x + (\cosh^2 y - 1) = 1$
$\sin^2 x + \cosh^2 y = 2$

(B) Given,$\tanh x = \frac{1}{2}$.
We know that $\tanh^2 x = 1 - \text{sech}^2 x$,so $\text{sech}^2 x = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Thus,$\text{sech } x = \frac{\sqrt{3}}{2}$ and $\cosh x = \frac{2}{\sqrt{3}}$.
Since $\tanh x = \frac{\sinh x}{\cosh x}$,we have $\sinh x = \tanh x \cdot \cosh x = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
Now,$\sinh 2x = 2 \sinh x \cosh x = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = \frac{4}{3}$.
And $\text{sech } 2x = \frac{1}{\cosh 2x} = \frac{1}{\cosh^2 x + \sinh^2 x} = \frac{1}{(\frac{2}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2} = \frac{1}{\frac{4}{3} + \frac{1}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}$.
Therefore,$\sinh 2x - \text{sech } 2x = \frac{4}{3} - \frac{3}{5} = \frac{20 - 9}{15} = \frac{11}{15}$.

(A) Given expression: $E = \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.
We know that $\sinh x = \frac{e^x - e^{-x}}{2}$,so $(e^x - e^{-x})^2 = 4\sinh^2 x$.
Thus,the denominator is $4(4\sinh^2 x) = 16\sinh^2 x$.
Alternatively,let us evaluate option $A$: $\sinh^2 x + \coth^2 x$.
$\sinh^2 x + \coth^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 + \left(\frac{e^x + e^{-x}}{e^x - e^{-x}}\right)^2$
$= \frac{(e^x - e^{-x})^4 + 4(e^x + e^{-x})^2}{4(e^x - e^{-x})^2}$
$= \frac{(e^{2x} + e^{-2x} - 2)^2 + 4(e^{2x} + e^{-2x} + 2)}{4(e^x - e^{-x})^2}$
$= \frac{(e^{4x} + e^{-4x} + 2 - 4e^{2x} - 4e^{-2x} + 2) + 4e^{2x} + 4e^{-2x} + 8}{4(e^x - e^{-x})^2}$
$= \frac{e^{4x} + e^{-4x} + 14}{4(e^x - e^{-x})^2}$.
This matches the given expression. Hence,the correct option is $A$.

(C) Given $|\sin \alpha - \cos \alpha| = \frac{3}{4}$.
Squaring both sides,we get $(\sin \alpha - \cos \alpha)^2 = (\frac{3}{4})^2$.
$\sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha = \frac{9}{16}$.
$1 - \sin 2\alpha = \frac{9}{16}$,which implies $\sin 2\alpha = 1 - \frac{9}{16} = \frac{7}{16}$.
Since $\sin 2\alpha = \frac{7}{16}$,we have the perpendicular $P = 7$ and hypotenuse $H = 16$.
The base $B = \sqrt{H^2 - P^2} = \sqrt{16^2 - 7^2} = \sqrt{256 - 49} = \sqrt{207} = 3\sqrt{23}$.
Thus,$\cos 2\alpha = \frac{B}{H} = \frac{3\sqrt{23}}{16}$.
Now,$|\sec 2\alpha - \tan 2\alpha| = |\frac{1}{\cos 2\alpha} - \frac{\sin 2\alpha}{\cos 2\alpha}| = |\frac{1 - \sin 2\alpha}{\cos 2\alpha}|$.
Substituting the values,$|\frac{1 - 7/16}{3\sqrt{23}/16}| = |\frac{9/16}{3\sqrt{23}/16}| = \frac{9}{3\sqrt{23}} = \frac{3}{\sqrt{23}}$.
Therefore,option $C$ is correct.

(A) Given $\sinh x = \tan A$.
We know that $\sinh x = \frac{e^x - e^{-x}}{2} = \tan A$,so $e^x - e^{-x} = 2 \tan A$.
Let $e^x = t$. Then $t - \frac{1}{t} = 2 \tan A$,which implies $t^2 - 2 \tan A \cdot t - 1 = 0$.
Solving for $t$ using the quadratic formula: $t = \frac{2 \tan A \pm \sqrt{4 \tan^2 A + 4}}{2} = \tan A \pm \sec A$.
Since $e^x > 0$,we take $e^x = \tan A + \sec A$.
Then $e^{-x} = \frac{1}{\sec A + \tan A} = \sec A - \tan A$.
Now,$\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{(\tan A + \sec A) - (\sec A - \tan A)}{(\tan A + \sec A) + (\sec A - \tan A)} = \frac{2 \tan A}{2 \sec A} = \frac{\tan A}{\sec A} = \sin A$.
Thus,$|\tanh x| = |\sin A|$.

(A) Given expression: $(1-\cos \theta)(1+\cos \theta)(1+\cot^2 \theta)$
Using the identity $(1-x)(1+x) = 1-x^2$,we get:
$(1-\cos^2 \theta)(1+\cot^2 \theta)$
Using the trigonometric identities $1-\cos^2 \theta = \sin^2 \theta$ and $1+\cot^2 \theta = \operatorname{cosec}^2 \theta$:
$= \sin^2 \theta \cdot \operatorname{cosec}^2 \theta$
$= \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = 1$
Since the expression simplifies to $1$ regardless of the value of $\theta$ (provided $\sin \theta \neq 0$),for $\theta = \frac{\pi}{15}$,the value is $1$.

(A) Let $S = \sum_{k=1}^{8} \cos^4 \frac{k\pi}{8}$.
Using the property $\cos(\pi - \theta) = -\cos \theta$,we have $\cos^4(\pi - \theta) = \cos^4 \theta$.
Thus,$\cos^4 \frac{7\pi}{8} = \cos^4 \frac{\pi}{8}$,$\cos^4 \frac{6\pi}{8} = \cos^4 \frac{2\pi}{8}$,and $\cos^4 \frac{5\pi}{8} = \cos^4 \frac{3\pi}{8}$.
Also,$\cos \frac{4\pi}{8} = \cos \frac{\pi}{2} = 0$ and $\cos \frac{8\pi}{8} = \cos \pi = -1$.
So,$S = 2(\cos^4 \frac{\pi}{8} + \cos^4 \frac{2\pi}{8} + \cos^4 \frac{3\pi}{8}) + 0^4 + (-1)^4$.
$S = 2(\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8} + (\frac{1}{\sqrt{2}})^4) + 1$.
Using $\cos^4 \theta + \sin^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta = 1 - \frac{1}{2}\sin^2(2\theta)$:
$S = 2(1 - \frac{1}{2}\sin^2 \frac{\pi}{4} + \frac{1}{4}) + 1 = 2(1 - \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{4}) + 1 = 2(1 - \frac{1}{4} + \frac{1}{4}) + 1 = 2(1) + 1 = 3$.

(B) Given equations are:
$\sin 2\theta + \sin 2\phi = \frac{1}{2}$ $(i)$
$\cos 2\theta + \cos 2\phi = \frac{3}{2}$ $(ii)$
Squaring both equations:
$(\sin 2\theta + \sin 2\phi)^2 = \frac{1}{4}$
$\sin^2 2\theta + \sin^2 2\phi + 2\sin 2\theta \sin 2\phi = \frac{1}{4}$ $(iii)$
$(\cos 2\theta + \cos 2\phi)^2 = \frac{9}{4}$
$\cos^2 2\theta + \cos^2 2\phi + 2\cos 2\theta \cos 2\phi = \frac{9}{4}$ $(iv)$
Adding $(iii)$ and $(iv)$:
$(\sin^2 2\theta + \cos^2 2\theta) + (\sin^2 2\phi + \cos^2 2\phi) + 2(\cos 2\theta \cos 2\phi + \sin 2\theta \sin 2\phi) = \frac{1}{4} + \frac{9}{4}$
$1 + 1 + 2\cos(2\theta - 2\phi) = \frac{10}{4}$
$2 + 2\cos 2(\theta - \phi) = \frac{5}{2}$
$2\cos 2(\theta - \phi) = \frac{5}{2} - 2 = \frac{1}{2}$
$\cos 2(\theta - \phi) = \frac{1}{4}$
Using the identity $\cos 2A = 2\cos^2 A - 1$:
$2\cos^2(\theta - \phi) - 1 = \frac{1}{4}$
$2\cos^2(\theta - \phi) = 1 + \frac{1}{4} = \frac{5}{4}$
$\cos^2(\theta - \phi) = \frac{5}{8}$

(D) Given equation is: $\cos \left(\frac{\pi}{4}-x\right) \cos 2 x+\sin x \sin 2 x \sec x = \cos x \sin 2 x \sec x+\cos \left(\frac{\pi}{4}+x\right) \cos 2 x$
Rearranging the terms: $\cos 2 x \left[ \cos \left(\frac{\pi}{4}-x\right) - \cos \left(\frac{\pi}{4}+x\right) \right] = \sin 2 x \sec x (\cos x - \sin x)$
Using the formula $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$:
$\cos 2 x \left[ 2 \sin \frac{\pi}{4} \sin x \right] = \sin 2 x \sec x (\cos x - \sin x)$
Since $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin 2 x = 2 \sin x \cos x$:
$\cos 2 x \left( \frac{2}{\sqrt{2}} \sin x \right) = (2 \sin x \cos x) \sec x (\cos x - \sin x)$
$\sqrt{2} \cos 2 x \sin x = 2 \sin x (\cos x - \sin x)$
Assuming $\sin x \neq 0$,we divide by $\sin x$:
$\sqrt{2} (\cos^2 x - \sin^2 x) = 2 (\cos x - \sin x)$
$\sqrt{2} (\cos x - \sin x)(\cos x + \sin x) = 2 (\cos x - \sin x)$
If $\cos x - \sin x \neq 0$,then $\sqrt{2} (\cos x + \sin x) = 2$,which implies $\cos x + \sin x = \sqrt{2}$.
Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = 1$,which is $\cos \left(x - \frac{\pi}{4}\right) = 1$.
Thus,$x - \frac{\pi}{4} = 0$,so $x = \frac{\pi}{4}$.
Then $\sec x = \sec \frac{\pi}{4} = \sqrt{2}$.

(B) Given,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ and $\beta = \frac{\cos^3 x}{\sin^2 x}$.
$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2 x} + \frac{\cos^4 x}{\sin^2 x} + 3$.
$= \frac{\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.
Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,where $a = \sin^2 x$ and $b = \cos^2 x$:
$= \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.
$= \frac{(\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.
Given $\sin x + \cos x = k$,squaring both sides gives $1 + 2 \sin x \cos x = k^2$,so $\sin x \cos x = \frac{k^2 - 1}{2}$.
Thus,$\frac{1}{\sin^2 x \cos^2 x} = \frac{1}{(\frac{k^2 - 1}{2})^2} = \frac{4}{(k^2 - 1)^2}$.

(C) Given equations are: $\sin \theta \cosh \alpha = \tan x$ and $\cos \theta \sinh \alpha = \sec x$.
Squaring and subtracting the equations: $\sec^2 x - \tan^2 x = (\cos \theta \sinh \alpha)^2 - (\sin \theta \cosh \alpha)^2$.
Since $\sec^2 x - \tan^2 x = 1$,we have: $\cos^2 \theta \sinh^2 \alpha - \sin^2 \theta \cosh^2 \alpha = 1$.
Using $\sin^2 \theta = 1 - \cos^2 \theta$ and $\sinh^2 \alpha = \cosh^2 \alpha - 1$:
$\cos^2 \theta (\cosh^2 \alpha - 1) - (1 - \cos^2 \theta) \cosh^2 \alpha = 1$.
$\cos^2 \theta \cosh^2 \alpha - \cos^2 \theta - \cosh^2 \alpha + \cos^2 \theta \cosh^2 \alpha = 1$.
$2 \cos^2 \theta \cosh^2 \alpha - \cos^2 \theta - \cosh^2 \alpha = 1$.
Using the identity $\cosh 2 \alpha = 2 \cosh^2 \alpha - 1 \implies \cosh^2 \alpha = \frac{\cosh 2 \alpha + 1}{2}$ and $\cos 2 \theta = 2 \cos^2 \theta - 1 \implies \cos^2 \theta = \frac{\cos 2 \theta + 1}{2}$.
Substituting these: $2 \left( \frac{\cos 2 \theta + 1}{2} \right) \left( \frac{\cosh 2 \alpha + 1}{2} \right) - \left( \frac{\cos 2 \theta + 1}{2} \right) - \left( \frac{\cosh 2 \alpha + 1}{2} \right) = 1$.
Multiplying by $2$: $(\cos 2 \theta + 1)(\cosh 2 \alpha + 1) - (\cos 2 \theta + 1) - (\cosh 2 \alpha + 1) = 2$.
$\cos 2 \theta \cosh 2 \alpha + \cos 2 \theta + \cosh 2 \alpha + 1 - \cos 2 \theta - 1 - \cosh 2 \alpha - 1 = 2$.
$\cos 2 \theta \cosh 2 \alpha - 1 = 2$.
Therefore,$\cos 2 \theta \cosh 2 \alpha = 3$.

(A) . The period of $\sin^2 x$ is $\pi$. Thus,$A-V$.
$B$. The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^2+b^2}$. Here,$\frac{\pi}{3} \sqrt{(\sqrt{3})^2+(1)^2} = \frac{\pi}{3} \times 2 = \frac{2\pi}{3}$. Thus,$B-I$.
$C$. The period of $\sin \frac{x}{3}$ is $\frac{2\pi}{1/3} = 6\pi$ and the period of $\cos \frac{x}{2}$ is $\frac{2\pi}{1/2} = 4\pi$. The period of the sum is the $LCM(6\pi, 4\pi) = 12\pi$. Thus,$C-II$.
$D$. For $y=|\sin x|$ and $y=1$,we have $|\sin x|=1$,so $\sin x = \pm 1$. In the interval $(0, \pi)$,$\sin x = 1$ at $x = \frac{\pi}{2}$. Thus,$D-III$.
Therefore,the correct match is $A-V, B-I, C-II, D-III$.

(B) Given,$\cos x - \sin x = \sqrt{a} \sin x$.
Rearranging the terms,we get $\cos x = \sin x(\sqrt{a} + 1)$.
Multiply both sides by $(\sqrt{a} - 1)$:
$(\sqrt{a} - 1) \cos x = \sin x(\sqrt{a} + 1)(\sqrt{a} - 1)$.
Using the identity $(x+y)(x-y) = x^2 - y^2$,we get:
$(\sqrt{a} - 1) \cos x = \sin x(a - 1)$.
$\sqrt{a} \cos x - \cos x = a \sin x - \sin x$.
Rearranging to solve for $a \sin x + \cos x - \sin x$:
$a \sin x + \cos x - \sin x = \sqrt{a} \cos x$.

(D) Let $S = \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} - \cos \frac{4\pi}{7} + \cos \frac{5\pi}{7} - \cos \frac{6\pi}{7}$.
Using the property $\cos(\pi - \theta) = -\cos \theta$,we have $\cos \frac{6\pi}{7} = -\cos \frac{\pi}{7}$,$\cos \frac{5\pi}{7} = -\cos \frac{2\pi}{7}$,and $\cos \frac{4\pi}{7} = -\cos \frac{3\pi}{7}$.
Substituting these into the expression:
$S = \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} - (-\cos \frac{3\pi}{7}) + (-\cos \frac{2\pi}{7}) - (-\cos \frac{\pi}{7})$
$S = \cos \frac{\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{3\pi}{7} - \cos \frac{2\pi}{7} + \cos \frac{\pi}{7}$
$S = 2 \cos \frac{\pi}{7} - 2 \cos \frac{2\pi}{7} + 2 \cos \frac{3\pi}{7}$.
Multiply and divide by $2 \sin \frac{\pi}{7}$:
$S = \frac{2 \sin \frac{\pi}{7} \cos \frac{\pi}{7} - 2 \sin \frac{\pi}{7} \cos \frac{2\pi}{7} + 2 \sin \frac{\pi}{7} \cos \frac{3\pi}{7}}{\sin \frac{\pi}{7}}$
Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:
$S = \frac{\sin \frac{2\pi}{7} - (\sin \frac{3\pi}{7} - \sin \frac{\pi}{7}) + (\sin \frac{4\pi}{7} - \sin \frac{2\pi}{7})}{\sin \frac{\pi}{7}}$
$S = \frac{\sin \frac{2\pi}{7} - \sin \frac{3\pi}{7} + \sin \frac{\pi}{7} + \sin \frac{4\pi}{7} - \sin \frac{2\pi}{7}}{\sin \frac{\pi}{7}}$
Since $\sin \frac{4\pi}{7} = \sin(\pi - \frac{4\pi}{7}) = \sin \frac{3\pi}{7}$,the terms cancel out:
$S = \frac{\sin \frac{\pi}{7}}{\sin \frac{\pi}{7}} = 1$.

(B) Given: $\sin x + \sin y = p$ $(i)$ and $\cos x + \cos y = q$ (ii).
Squaring and adding $(i)$ and (ii):
$(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = p^2 + q^2$
$(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = p^2 + q^2$
$1 + 1 + 2 \cos(x - y) = p^2 + q^2 \implies 2 + 2 \cos(x - y) = p^2 + q^2$ (iii).
Now,consider $q^2 - p^2 = (\cos x + \cos y)^2 - (\sin x + \sin y)^2$
$= (\cos^2 x - \sin^2 x) + (\cos^2 y - \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$
$= \cos 2x + \cos 2y + 2 \cos(x + y)$
$= 2 \cos(x + y) \cos(x - y) + 2 \cos(x + y) = 2 \cos(x + y) [\cos(x - y) + 1]$.
From (iii),$2 \cos(x - y) + 2 = p^2 + q^2$,so $\cos(x - y) + 1 = \frac{p^2 + q^2}{2}$.
Substituting this into the expression for $q^2 - p^2$:
$q^2 - p^2 = 2 \cos(x + y) \cdot \frac{p^2 + q^2}{2} = \cos(x + y)(p^2 + q^2)$.
Therefore,$\sec(x + y) = \frac{1}{\cos(x + y)} = \frac{p^2 + q^2}{q^2 - p^2}$.
Thus,option $B$ is correct.

(C) Given: $\cosh \beta = \sec \alpha \cos \theta$ and $\sinh \beta = \operatorname{cosec} \alpha \sin \theta$.
From the second equation,$\sin \theta = \sinh \beta \sin \alpha$.
Squaring both sides,$\sin^2 \theta = \sinh^2 \beta \sin^2 \alpha$.
We know that $\cosh^2 \beta - \sinh^2 \beta = 1$,so $\cosh^2 \beta = 1 + \sinh^2 \beta$.
Substitute $\cosh \beta = \sec \alpha \cos \theta$ into the identity:
$\sec^2 \alpha \cos^2 \theta = 1 + \sinh^2 \beta$.
$\sec^2 \alpha (1 - \sin^2 \theta) = 1 + \sinh^2 \beta$.
Substitute $\sin^2 \theta = \sinh^2 \beta \sin^2 \alpha$:
$\sec^2 \alpha (1 - \sinh^2 \beta \sin^2 \alpha) = 1 + \sinh^2 \beta$.
$\sec^2 \alpha - \sec^2 \alpha \sin^2 \alpha \sinh^2 \beta = 1 + \sinh^2 \beta$.
Since $\sec^2 \alpha \sin^2 \alpha = \tan^2 \alpha$:
$\sec^2 \alpha - \tan^2 \alpha \sinh^2 \beta = 1 + \sinh^2 \beta$.
$\sec^2 \alpha - 1 = \sinh^2 \beta (1 + \tan^2 \alpha)$.
$\tan^2 \alpha = \sinh^2 \beta \sec^2 \alpha$.
$\sinh^2 \beta = \frac{\tan^2 \alpha}{\sec^2 \alpha} = \sin^2 \alpha$.

(B) First,calculate $m = \sin 10^{\circ} \sin 50^{\circ} \sin 60^{\circ} \sin 70^{\circ}$.
Using the identity $\sin \theta \sin(60^{\circ}-\theta) \sin(60^{\circ}+\theta) = \frac{1}{4} \sin 3\theta$,we have $\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ} = \frac{1}{4} \sin(3 \times 10^{\circ}) = \frac{1}{4} \sin 30^{\circ} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$.
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,then $m = \frac{1}{8} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{16}$.
Next,calculate $n = \tan 20^{\circ} \tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}$.
Using the identity $\tan \theta \tan(60^{\circ}-\theta) \tan(60^{\circ}+\theta) = \tan 3\theta$,we have $\tan 20^{\circ} \tan 40^{\circ} \tan 80^{\circ} = \tan(3 \times 20^{\circ}) = \tan 60^{\circ} = \sqrt{3}$.
Since $\tan 60^{\circ} = \sqrt{3}$,then $n = \sqrt{3} \times \sqrt{3} = 3$.
Finally,$\frac{n}{m} = \frac{3}{\frac{\sqrt{3}}{16}} = \frac{3 \times 16}{\sqrt{3}} = 16 \sqrt{3}$.

(C) Given,$\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$.
Squaring both sides:
$\left(\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n}\right)^2 = \frac{n}{4}$
$1 + \sin \frac{\pi}{n} = \frac{n}{4}$
$\sin \frac{\pi}{n} = \frac{n-4}{4}$.
Since $n > 2$,$\frac{\pi}{n} \in (0, \frac{\pi}{2}]$,so $\sin \frac{\pi}{n} > 0$,which implies $n-4 > 0$,so $n > 4$.
Also,$\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \sqrt{2} \sin \left(\frac{\pi}{4} + \frac{\pi}{2n}\right) = \frac{\sqrt{n}}{2}$.
$\sin \left(\frac{\pi}{4} + \frac{\pi}{2n}\right) = \frac{\sqrt{n}}{2\sqrt{2}} = \sqrt{\frac{n}{8}}$.
Since $\sin \theta \le 1$,we have $\sqrt{\frac{n}{8}} \le 1 \Rightarrow n \le 8$.
For $n=8$,$\sin \frac{\pi}{16} = \frac{8-4}{4} = 1$,which is impossible as $\frac{\pi}{16} \neq \frac{\pi}{2}$.
Thus,$4 < n < 8$,so $n \in \{5, 6, 7\}$.
The number of integral values is $3$.

(C) Given the inequality $|\sqrt{3} \cos x - \sin x| \geq 2$.
We know that the maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$.
Here,$a = \sqrt{3}$ and $b = -1$,so $\sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$.
Thus,the expression $|\sqrt{3} \cos x - \sin x|$ can only be $\geq 2$ when it is exactly $2$.
This implies $2 \cos(x + \frac{\pi}{6}) = 2$ or $2 \cos(x + \frac{\pi}{6}) = -2$.
$\cos(x + \frac{\pi}{6}) = 1 \implies x + \frac{\pi}{6} = 0, 2\pi \implies x = \frac{11\pi}{6}$ (within $[0, 2\pi]$).
$\cos(x + \frac{\pi}{6}) = -1 \implies x + \frac{\pi}{6} = \pi \implies x = \frac{5\pi}{6}$.
So,$A = \{\frac{5\pi}{6}, \frac{11\pi}{6}\}$.
Taking $x_1 = \frac{5\pi}{6}$ and $x_2 = \frac{11\pi}{6}$,we get $\frac{x_1}{x_2} = \frac{5\pi/6}{11\pi/6} = \frac{5}{11}$.

(D) Given,$\cos x+\cos y=-\cos \alpha$ $(i)$
Using the formula $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get:
$2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\cos \alpha$
Also,$\sin x+\sin y=-\sin \alpha$ (ii)
Using the formula $\sin C + \sin D = 2 \sin \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$,we get:
$2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right) = -\sin \alpha$
Dividing equation $(i)$ by equation (ii):
$\frac{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)} = \frac{-\cos \alpha}{-\sin \alpha}$
$\cot \left(\frac{x+y}{2}\right) = \cot \alpha$

(C) Given expression: $\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$
Divide the numerator and denominator of the first term by $\cos 13^{\circ}$:
$= \frac{1-\tan 13^{\circ}}{1+\tan 13^{\circ}} + \tan 148^{\circ}$
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}$,where $A=45^{\circ}$ and $B=13^{\circ}$:
$= \tan(45^{\circ}-13^{\circ}) + \tan(180^{\circ}-32^{\circ})$
$= \tan 32^{\circ} - \tan 32^{\circ} = 0$

(B) Given: $a \sin^2 \theta + b \cos^2 \theta = c$
Divide both sides by $\cos^2 \theta$:
$a \tan^2 \theta + b = c \sec^2 \theta$
Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$:
$a \tan^2 \theta + b = c(1 + \tan^2 \theta)$
$a \tan^2 \theta + b = c + c \tan^2 \theta$
Rearranging the terms to isolate $\tan^2 \theta$:
$a \tan^2 \theta - c \tan^2 \theta = c - b$
$(a - c) \tan^2 \theta = c - b$
$\tan^2 \theta = \frac{c - b}{a - c}$

(D) We have,$\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$.
Using the sum-to-product formula $\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}$:
$= (\cos 132^{\circ} + \cos 12^{\circ}) + (\cos 156^{\circ} + \cos 84^{\circ})$
$= 2 \cos \frac{132^{\circ}+12^{\circ}}{2} \cos \frac{132^{\circ}-12^{\circ}}{2} + 2 \cos \frac{156^{\circ}+84^{\circ}}{2} \cos \frac{156^{\circ}-84^{\circ}}{2}$
$= 2 \cos 72^{\circ} \cos 60^{\circ} + 2 \cos 120^{\circ} \cos 36^{\circ}$
$= 2 \left( \frac{\sqrt{5}-1}{4} \right) \left( \frac{1}{2} \right) + 2 \left( -\frac{1}{2} \right) \left( \frac{\sqrt{5}+1}{4} \right)$
$= \frac{\sqrt{5}-1}{4} - \frac{\sqrt{5}+1}{4}$
$= \frac{\sqrt{5}-1-\sqrt{5}-1}{4} = -\frac{2}{4} = -\frac{1}{2}$.

(A) We know that $\tan 63^{\circ} = \cot 27^{\circ}$ and $\tan 81^{\circ} = \cot 9^{\circ}$.
Substituting these into the expression:
$\tan 9^{\circ} - \tan 27^{\circ} - \cot 27^{\circ} + \cot 9^{\circ} = (\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$
Using the identity $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}$:
$= \frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$
Given $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$:
$= \frac{2}{(\sqrt{5}-1)/4} - \frac{2}{(\sqrt{5}+1)/4} = \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1}$
$= 8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \right) = 8 \left( \frac{2}{5-1} \right) = 8 \left( \frac{2}{4} \right) = 4$.

(C) We have $\sum_{k=1}^3 \cos ^2\left((2 k-1) \frac{\pi}{12}\right) = \cos ^2 \frac{\pi}{12} + \cos ^2 \frac{3 \pi}{12} + \cos ^2 \frac{5 \pi}{12}$.
Since $\frac{3 \pi}{12} = \frac{\pi}{4}$,we have $\cos ^2 \frac{\pi}{4} = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
Thus,the expression becomes $\cos ^2 \frac{\pi}{12} + \frac{1}{2} + \cos ^2 \frac{5 \pi}{12}$.
Using the identity $\cos(\frac{\pi}{2} - \theta) = \sin \theta$,we have $\cos \frac{5 \pi}{12} = \cos(\frac{\pi}{2} - \frac{\pi}{12}) = \sin \frac{\pi}{12}$.
Substituting this,we get $\frac{1}{2} + \cos ^2 \frac{\pi}{12} + \sin ^2 \frac{\pi}{12}$.
Since $\cos ^2 \theta + \sin ^2 \theta = 1$,the expression simplifies to $\frac{1}{2} + 1 = \frac{3}{2}$.

(A) Given $\operatorname{cosec} \theta = \frac{p+q}{p-q}$.
We know that $\sin \theta = \frac{p-q}{p+q}$.
Using the formula $\sin \theta = \frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)}$,we have:
$\frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{p-q}{p+q}$
Applying componendo and dividendo:
$\frac{1 + \tan^2(\theta/2) + 2 \tan(\theta/2)}{1 + \tan^2(\theta/2) - 2 \tan(\theta/2)} = \frac{(p+q) + (p-q)}{(p+q) - (p-q)}$
$\frac{(1 + \tan(\theta/2))^2}{(1 - \tan(\theta/2))^2} = \frac{2p}{2q} = \frac{p}{q}$
Taking the square root on both sides:
$\frac{1 + \tan(\theta/2)}{1 - \tan(\theta/2)} = \sqrt{\frac{p}{q}}$
Since $\tan(\pi/4) = 1$,the expression becomes:
$\tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \sqrt{\frac{p}{q}}$
Therefore,$\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1}{\tan(\pi/4 + \theta/2)} = \sqrt{\frac{q}{p}}$.

(A) Given the inequality: $|\sin x-\cos ^2 x| \geq|3-3 \sin x+\sin ^2 x|+4|\sin x-1|$
Substitute $\cos ^2 x = 1-\sin ^2 x$:
$|\sin x-(1-\sin ^2 x)| \geq|\sin ^2 x-3 \sin x+3|+4|\sin x-1|$
$|\sin ^2 x+\sin x-1| \geq|\sin ^2 x-3 \sin x+3|+|4 \sin x-4|$
Note that $(\sin ^2 x-3 \sin x+3) + (4 \sin x-4) = \sin ^2 x+\sin x-1$.
Let $a = \sin ^2 x-3 \sin x+3$ and $b = 4 \sin x-4$. The inequality is $|a+b| \geq |a|+|b|$.
This holds if and only if $ab \geq 0$.
Since the discriminant of $a = \sin ^2 x-3 \sin x+3$ is $D = (-3)^2 - 4(1)(3) = 9-12 = -3 < 0$ and the coefficient of $\sin ^2 x$ is positive,$a > 0$ for all $x \in R$.
Thus,$b \geq 0$ $\Rightarrow 4(\sin x-1) \geq 0$ $\Rightarrow \sin x \geq 1$.
Since $\sin x \leq 1$,we must have $\sin x = 1$.
Therefore,$x = 2n\pi + \frac{\pi}{2} = (4n+1)\frac{\pi}{2}, n \in Z$.

(C) Given $\sinh x = -\frac{1}{2}$.
We know the identity $\cosh^2 x - \sinh^2 x = 1$.
Substituting the value: $\cosh^2 x = 1 + (-\frac{1}{2})^2 = 1 + \frac{1}{4} = \frac{5}{4}$.
Since $\cosh x > 0$,we have $\cosh x = \frac{\sqrt{5}}{2}$.
Now,$\tanh x = \frac{\sinh x}{\cosh x} = \frac{-1/2}{\sqrt{5}/2} = -\frac{1}{\sqrt{5}}$.
Using the double angle formula $\tanh 2x = \frac{2 \tanh x}{1 + \tanh^2 x}$:
$\tanh 2x = \frac{2(-1/\sqrt{5})}{1 + (-1/\sqrt{5})^2} = \frac{-2/\sqrt{5}}{1 + 1/5} = \frac{-2/\sqrt{5}}{6/5} = -\frac{2}{\sqrt{5}} \times \frac{5}{6} = -\frac{\sqrt{5}}{3}$.

(B) Let $a$ be the side length and $b$ be the diagonal length of a regular pentagon inscribed in a circle.
In a regular pentagon,the interior angle is $\frac{3\pi}{5}$.
Consider the triangle formed by two sides and a diagonal. By properties of a regular pentagon,the diagonal $b$ subtends an angle of $\frac{2\pi}{5}$ at the center,and the angle between a side and a diagonal is $\frac{\pi}{5}$.
Drawing a perpendicular from vertex $C$ to the diagonal $AB$ at point $D$,we form a right-angled triangle $\triangle ACD$.
In $\triangle ACD$,the angle $\angle CAD = \frac{\pi}{5}$ and the hypotenuse $AC = a$.
Thus,$\cos \frac{\pi}{5} = \frac{AD}{AC} = \frac{AD}{a}$,which implies $AD = a \cos \frac{\pi}{5}$.
Since the diagonal $b$ is bisected by the perpendicular from the opposite vertex in a regular pentagon,$b = 2AD$.
Therefore,$b = 2a \cos \frac{\pi}{5}$,which gives $\frac{b}{a} = 2 \cos \frac{\pi}{5}$.

(A) Given $\operatorname{Sinh}^{-1} x = \log(1+\sqrt{2})$.
Using the definition $\operatorname{Sinh}^{-1} x = \log(x + \sqrt{x^2+1})$,we have $x + \sqrt{x^2+1} = 1+\sqrt{2}$.
Comparing the terms,we get $x = 1$.
Given $\operatorname{Cosh}^{-1} y = \log(1+\sqrt{2})$.
Using the definition $\operatorname{Cosh}^{-1} y = \log(y + \sqrt{y^2-1})$,we have $y + \sqrt{y^2-1} = 1+\sqrt{2}$.
This implies $y = \sqrt{2}$.
Now,we need to find $\operatorname{Tan}^{-1}(x+y) = \operatorname{Tan}^{-1}(1+\sqrt{2})$.
We know that $\tan(67.5^{\circ}) = \tan(\frac{135^{\circ}}{2}) = \frac{1-\cos(135^{\circ})}{\sin(135^{\circ})} = \frac{1 - (-1/\sqrt{2})}{1/\sqrt{2}} = \sqrt{2}+1$.
Therefore,$\operatorname{Tan}^{-1}(1+\sqrt{2}) = 67.5^{\circ} = 67 \frac{1}{2}^{\circ}$.

(D) Let $y = \operatorname{sech}^{-1}(\sin \theta)$.
Then,$\operatorname{sech} y = \sin \theta$.
Since $\operatorname{sech} y = \frac{1}{\cosh y}$,we have $\cosh y = \frac{1}{\sin \theta} = \operatorname{cosec} \theta$.
Thus,$y = \cosh^{-1}(\operatorname{cosec} \theta)$.
Using the logarithmic form of the inverse hyperbolic cosine function,$\cosh^{-1}(x) = \log(x + \sqrt{x^2 - 1})$,we get:
$y = \log(\operatorname{cosec} \theta + \sqrt{\operatorname{cosec}^2 \theta - 1})$.
Since $\operatorname{cosec}^2 \theta - 1 = \cot^2 \theta$,we have:
$y = \log(\operatorname{cosec} \theta + \cot \theta)$.
Using the trigonometric identities $\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{1}{2 \sin(\theta/2) \cos(\theta/2)}$ and $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\cos^2(\theta/2) - \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}$,we get:
$y = \log\left(\frac{1 + \cos^2(\theta/2) - \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}\right) = \log\left(\frac{2 \cos^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}\right) = \log(\cot(\theta/2))$.
Therefore,$\operatorname{sech}^{-1}(\sin \theta) = \log \cot \frac{\theta}{2}$.

(B) We have,$\sinh ^{-1}(\sqrt{8})+\sinh ^{-1}(\sqrt{24})=\alpha$.
Let $\sinh ^{-1}(\sqrt{8})=x$,then $\sinh x = \sqrt{8}$.
Since $\cosh^2 x - \sinh^2 x = 1$,we have $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = \sqrt{9} = 3$.
Let $\sinh ^{-1}(\sqrt{24})=y$,then $\sinh y = \sqrt{24}$.
Similarly,$\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + 24} = \sqrt{25} = 5$.
Using the identity $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$,we get:
$\sinh(x+y) = (\sqrt{8})(5) + (3)(\sqrt{24}) = 5(2\sqrt{2}) + 3(2\sqrt{6}) = 10\sqrt{2} + 6\sqrt{6}$.
Since $\alpha = x+y$,we have $\sinh \alpha = 6\sqrt{6} + 10\sqrt{2}$.

(A) We know that the formula for the inverse hyperbolic sine function is $\sinh ^{-1}(y) = \log \left(y + \sqrt{1 + y^2}\right)$.
Substituting $y = \cot x$ into the formula,we get:
$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{1 + \cot ^2 x}\right)$
Since $1 + \cot ^2 x = \operatorname{cosec}^2 x$,we have:
$\sinh ^{-1}(\cot x) = \log \left(\cot x + \sqrt{\operatorname{cosec}^2 x}\right) = \log (\cot x + \operatorname{cosec} x)$
Using trigonometric identities $\cot x = \frac{\cos x}{\sin x}$ and $\operatorname{cosec} x = \frac{1}{\sin x}$,we get:
$\log \left(\frac{\cos x + 1}{\sin x}\right)$
Using half-angle formulas $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$\log \left(\frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) = \log \left(\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) = \log \left(\cot \frac{x}{2}\right)$.

(B) Given that,$x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$.
This implies $e^x = \cot \left( \frac{\pi}{4} + \theta \right)$ and $e^{-x} = \tan \left( \frac{\pi}{4} + \theta \right)$.
We know that $\sinh x = \frac{e^x - e^{-x}}{2}$.
Substituting the values,we get:
$\sinh x = \frac{\cot \left( \frac{\pi}{4} + \theta \right) - \tan \left( \frac{\pi}{4} + \theta \right)}{2}$
Using the identity $\cot A - \tan A = \frac{\cos A}{\sin A} - \frac{\sin A}{\cos A} = \frac{\cos^2 A - \sin^2 A}{\sin A \cos A} = \frac{\cos 2A}{\frac{1}{2} \sin 2A} = 2 \cot 2A$,we have:
$\sinh x = \frac{1}{2} \left[ 2 \cot \left( 2 \left( \frac{\pi}{4} + \theta \right) \right) \right]$
$\sinh x = \cot \left( \frac{\pi}{2} + 2\theta \right)$
Since $\cot \left( \frac{\pi}{2} + \alpha \right) = -\tan \alpha$,we get:
$\sinh x = -\tan 2\theta$.

(A) Given that,$e^{\sin x}-e^{-\sin x}-4=0$.
Let $e^{\sin x}=t$. Since $e^{\sin x} > 0$,we have $t > 0$.
The equation becomes $t - \frac{1}{t} - 4 = 0$,which simplifies to $t^2 - 4t - 1 = 0$.
Solving for $t$ using the quadratic formula: $t = \frac{4 \pm \sqrt{16 - 4(1)(-1)}}{2} = \frac{4 \pm \sqrt{20}}{2} = 2 \pm \sqrt{5}$.
Since $t > 0$,we must have $t = 2 + \sqrt{5}$ (as $2 - \sqrt{5} < 0$).
Thus,$e^{\sin x} = 2 + \sqrt{5}$.
We know that $-1 \leq \sin x \leq 1$,which implies $e^{-1} \leq e^{\sin x} \leq e^1$.
Numerically,$e \approx 2.718$ and $e^{-1} \approx 0.368$.
Also,$2 + \sqrt{5} \approx 2 + 2.236 = 4.236$.
Since $4.236 > 2.718$,the equation $e^{\sin x} = 2 + \sqrt{5}$ has no solution for $x$.
Therefore,the number of real values of $x$ is $0$.

(A) Given,$0 \leq P, Q \leq \frac{\pi}{2}$ and $\sin P + \cos Q = 2.$
Since the maximum value of $\sin P$ is $1$ and the maximum value of $\cos Q$ is $1,$ the equation $\sin P + \cos Q = 2$ holds only when $\sin P = 1$ and $\cos Q = 1.$
For $0 \leq P \leq \frac{\pi}{2},$ $\sin P = 1$ implies $P = \frac{\pi}{2}.$
For $0 \leq Q \leq \frac{\pi}{2},$ $\cos Q = 1$ implies $Q = 0.$
Therefore,$\tan \left(\frac{P + Q}{2}\right) = \tan \left(\frac{\frac{\pi}{2} + 0}{2}\right) = \tan \left(\frac{\pi}{4}\right) = 1.$

(D) Given,$\sin ^{2} \theta+3 \cos \theta=2$
$\Rightarrow 1-\cos ^{2} \theta+3 \cos \theta=2$
$\Rightarrow \cos ^{2} \theta-3 \cos \theta+1=0$
Using the quadratic formula,$\cos \theta = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$
Since $-1 \leq \cos \theta \leq 1$ and $\frac{3+\sqrt{5}}{2} > 1$,we must have $\cos \theta = \frac{3-\sqrt{5}}{2}$.
Then $\sec \theta = \frac{1}{\cos \theta} = \frac{2}{3-\sqrt{5}} = \frac{2(3+\sqrt{5})}{9-5} = \frac{3+\sqrt{5}}{2}$.
Now,$\cos \theta + \sec \theta = \frac{3-\sqrt{5}}{2} + \frac{3+\sqrt{5}}{2} = \frac{6}{2} = 3$.
Also,$\cos \theta \cdot \sec \theta = 1$.
We know that $\cos ^{3} \theta+\sec ^{3} \theta = (\cos \theta+\sec \theta)^{3} - 3 \cos \theta \sec \theta (\cos \theta+\sec \theta)$.
Substituting the values: $(3)^{3} - 3(1)(3) = 27 - 9 = 18$.

(C) Given $\sin \theta = \frac{2t}{1+t^2}$.
We know that $\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{2t}{1+t^2}\right)^2$.
$\cos^2 \theta = 1 - \frac{4t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 - 4t^2}{(1+t^2)^2} = \frac{1 + 2t^2 + t^4 - 4t^2}{(1+t^2)^2} = \frac{1 - 2t^2 + t^4}{(1+t^2)^2} = \frac{(1-t^2)^2}{(1+t^2)^2}$.
Taking the square root,$|\cos \theta| = \frac{|1-t^2|}{1+t^2}$.
Since $\theta$ lies in the second quadrant,$\cos \theta$ must be negative.
Therefore,$\cos \theta = -\frac{|1-t^2|}{1+t^2}$.