Properties of ITF Questions in English

(A) Let $x^2 = \cos 2\theta$,which implies $\theta = \frac{1}{2} \cos^{-1} x^2$.
Substituting this into the expression:
$\tan^{-1} \left[ \frac{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}} \right]$
Using the identities $1+\cos 2\theta = 2\cos^2 \theta$ and $1-\cos 2\theta = 2\sin^2 \theta$:
$= \tan^{-1} \left[ \frac{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta} \right]$
Dividing numerator and denominator by $\sqrt{2}\cos \theta$:
$= \tan^{-1} \left[ \frac{1 + \tan \theta}{1 - \tan \theta} \right]$
Using the formula $\tan(\frac{\pi}{4} + \theta) = \frac{\tan \frac{\pi}{4} + \tan \theta}{1 - \tan \frac{\pi}{4} \tan \theta}$:
$= \tan^{-1} \left[ \tan \left( \frac{\pi}{4} + \theta \right) \right] = \frac{\pi}{4} + \theta$
Substituting back $\theta = \frac{1}{2} \cos^{-1} x^2$:
$= \frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2$.

(B) Given the equation: $\sin^{-1}x - \cos^{-1}x = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
We know that $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.
So,the equation becomes: $\sin^{-1}x - \cos^{-1}x = \frac{\pi}{6}$.
We also know the identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.
Adding these two equations:
$(\sin^{-1}x - \cos^{-1}x) + (\sin^{-1}x + \cos^{-1}x) = \frac{\pi}{6} + \frac{\pi}{2}$.
$2\sin^{-1}x = \frac{\pi + 3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.
$\sin^{-1}x = \frac{\pi}{3}$.
$x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
Since we found exactly one value for $x$,the equation has a unique solution.

(A) We have $\sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{2m}}{{{m^4} + {m^2} + 2}}} \right)} $
$= \sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{2m}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} \right)} $
$= \sum\limits_{m = 1}^n {{{\tan }^{ - 1}}\left( {\frac{{({m^2} + m + 1) - ({m^2} - m + 1)}}{{1 + ({m^2} + m + 1)({m^2} - m + 1)}}} \right)} $
$= \sum\limits_{m = 1}^n {[{{\tan }^{ - 1}}({m^2} + m + 1) - {{\tan }^{ - 1}}({m^2} - m + 1)]} $
$= ({\tan ^{ - 1}}3 - {\tan ^{ - 1}}1) + ({\tan ^{ - 1}}7 - {\tan ^{ - 1}}3) + ({\tan ^{ - 1}}13 - {\tan ^{ - 1}}7) + \dots + [{\tan ^{ - 1}}({n^2} + n + 1) - {\tan ^{ - 1}}({n^2} - n + 1)]$
$= {\tan ^{ - 1}}({n^2} + n + 1) - {\tan ^{ - 1}}1$
$= {\tan ^{ - 1}}\left( {\frac{{({n^2} + n + 1) - 1}}{{1 + ({n^2} + n + 1)(1)}}} \right) = {\tan ^{ - 1}}\left( {\frac{{{n^2} + n}}{{{n^2} + n + 2}}} \right)$.

(A) Given equation is $2\cos^{-1}x + \sin^{-1}x = \frac{11\pi}{6}$.
We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Rewriting the given equation: $\cos^{-1}x + (\cos^{-1}x + \sin^{-1}x) = \frac{11\pi}{6}$.
Substituting the identity: $\cos^{-1}x + \frac{\pi}{2} = \frac{11\pi}{6}$.
Solving for $\cos^{-1}x$: $\cos^{-1}x = \frac{11\pi}{6} - \frac{\pi}{2} = \frac{11\pi - 3\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3}$.
Since the range of $\cos^{-1}x$ is $[0, \pi]$,and $\frac{4\pi}{3} > \pi$,there is no value of $x$ that satisfies this equation.
Therefore,the equation has no solution.

(A) Given the equation: $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$.
Using the formula for the sum of three inverse tangents: $\tan^{-1}\left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right) = \pi$.
Taking the tangent of both sides: $\frac{x+y+z-xyz}{1-xy-yz-zx} = \tan(\pi)$.
Since $\tan(\pi) = 0$,we have: $\frac{x+y+z-xyz}{1-xy-yz-zx} = 0$.
This implies the numerator must be zero: $x+y+z-xyz = 0$.
Therefore: $x+y+z = xyz$.

(A) We use the formula $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$.
Let $x = \sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}$.
Then $2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right] = {\cos ^{ - 1}}\left[ {\frac{{1 - \left( {\frac{{a - b}}{{a + b}}} \right){{\tan }^2}\frac{\theta }{2}}}{{1 + \left( {\frac{{a - b}}{{a + b}}} \right){{\tan }^2}\frac{\theta }{2}}}} \right]$
$= {\cos ^{ - 1}}\left[ {\frac{{(a + b) - (a - b){{\tan }^2}\frac{\theta }{2}}}{{(a + b) + (a - b){{\tan }^2}\frac{\theta }{2}}}} \right]$
$= {\cos ^{ - 1}}\left[ {\frac{{a(1 - {{\tan }^2}\frac{\theta }{2}) + b(1 + {{\tan }^2}\frac{\theta }{2})}}{{a(1 + {{\tan }^2}\frac{\theta }{2}) + b(1 - {{\tan }^2}\frac{\theta }{2})}}} \right]$
Dividing numerator and denominator by $(1 + {\tan ^2}\frac{\theta }{2})$:
$= {\cos ^{ - 1}}\left[ {\frac{{a\left( {\frac{{1 - {{\tan }^2}\frac{\theta }{2}}}{{1 + {{\tan }^2}\frac{\theta }{2}}}} \right) + b}}{{a + b\left( {\frac{{1 - {{\tan }^2}\frac{\theta }{2}}}{{1 + {{\tan }^2}\frac{\theta }{2}}}} \right)}}} \right]$
Using the identity $\cos \theta = \frac{{1 - {{\tan }^2}\frac{\theta }{2}}}{{1 + {{\tan }^2}\frac{\theta }{2}}}$,we get:
$= {\cos ^{ - 1}}\left( {\frac{{a\cos \theta + b}}{{a + b\cos \theta }}} \right)$.

(D) Given the equation: $2 \tan^{-1}(\cos x) = \tan^{-1}(\csc^2 x)$.
Using the formula $2 \tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,we get:
$\tan^{-1}\left(\frac{2 \cos x}{1 - \cos^2 x}\right) = \tan^{-1}(\csc^2 x)$.
Since $1 - \cos^2 x = \sin^2 x$,the equation becomes:
$\frac{2 \cos x}{\sin^2 x} = \frac{1}{\sin^2 x}$.
Assuming $\sin^2 x \neq 0$,we can cancel $\sin^2 x$ from both sides:
$2 \cos x = 1$.
$\cos x = \frac{1}{2}$.
Thus,$x = \frac{\pi}{3}$.

(C) Let $x = \sin A$ and $\sqrt{x} = \sin B$.
Then $A = \sin^{-1} x$ and $B = \sin^{-1} \sqrt{x}$.
The expression becomes $y = \sin^{-1}(\sin A \cos B + \sin B \cos A)$.
Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get $y = \sin^{-1}(\sin(A + B)) = A + B$.
Substituting back,$y = \sin^{-1} x + \sin^{-1} \sqrt{x}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} \sqrt{x})$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}$.
$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2\sqrt{x - x^2}}$.

(C) Given $y = \tan^{-1} \left( \frac{4x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2 + 3x}{3 - 2x} \right)$.
First,simplify the first term: $\tan^{-1} \left( \frac{5x - x}{1 + 5x \cdot x} \right) = \tan^{-1}(5x) - \tan^{-1}(x)$.
Next,simplify the second term: $\tan^{-1} \left( \frac{\frac{2}{3} + x}{1 - \frac{2}{3} \cdot x} \right) = \tan^{-1} \left( \frac{2}{3} \right) + \tan^{-1}(x)$.
Adding these together,$y = \tan^{-1}(5x) - \tan^{-1}(x) + \tan^{-1} \left( \frac{2}{3} \right) + \tan^{-1}(x) = \tan^{-1}(5x) + \tan^{-1} \left( \frac{2}{3} \right)$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\tan^{-1}(5x)] + \frac{d}{dx} [\tan^{-1} \left( \frac{2}{3} \right)]$.
Since $\tan^{-1} \left( \frac{2}{3} \right)$ is a constant,its derivative is $0$.
Therefore,$\frac{dy}{dx} = \frac{1}{1 + (5x)^2} \cdot \frac{d}{dx}(5x) = \frac{5}{1 + 25x^2}$.

(A) Given the expression: $y = \sec^{-1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) + \sin^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right)$
We know that $\sec^{-1}(z) = \cos^{-1}\left( \frac{1}{z} \right)$.
Therefore,$\sec^{-1}\left( \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \right) = \cos^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right)$.
Substituting this into the original equation:
$y = \cos^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right) + \sin^{-1}\left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right)$
Using the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$,we get:
$y = \frac{\pi}{2}$
Since $\frac{\pi}{2}$ is a constant,its derivative with respect to $x$ is:
$\frac{dy}{dx} = 0$

(A) Let $y = \sin^{-1}(3x - 4x^3)$.
Substitute $x = \sin \theta$,which implies $\theta = \sin^{-1} x$.
Then,$y = \sin^{-1}(3 \sin \theta - 4 \sin^3 \theta)$.
Using the trigonometric identity $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$,we get $y = \sin^{-1}(\sin 3\theta) = 3\theta$.
Substituting back $\theta = \sin^{-1} x$,we have $y = 3 \sin^{-1} x$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1} x) = 3 \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{3}{\sqrt{1 - x^2}}$.

(A) Given $y = \tan^{-1} \left( \frac{x^{1/3} + a^{1/3}}{1 - x^{1/3}a^{1/3}} \right)$.
Using the formula $\tan^{-1} \left( \frac{u + v}{1 - uv} \right) = \tan^{-1} u + \tan^{-1} v$,we get:
$y = \tan^{-1}(x^{1/3}) + \tan^{-1}(a^{1/3})$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} [\tan^{-1}(x^{1/3})] + \frac{d}{dx} [\tan^{-1}(a^{1/3})]$.
Since $a$ is a constant,$\frac{d}{dx} [\tan^{-1}(a^{1/3})] = 0$.
Using the chain rule,$\frac{d}{dx} [\tan^{-1}(x^{1/3})] = \frac{1}{1 + (x^{1/3})^2} \cdot \frac{d}{dx}(x^{1/3})$.
$= \frac{1}{1 + x^{2/3}} \cdot \frac{1}{3} x^{-2/3} = \frac{1}{3x^{2/3}(1 + x^{2/3})}$.
Thus,$\frac{dy}{dx} = \frac{1}{3x^{2/3}(1 + x^{2/3})}$.

(B) Given $y = \cot^{-1} \left( \frac{1 + x}{1 - x} \right)$.
We know that $\frac{1 + x}{1 - x} = \tan \left( \frac{\pi}{4} + \tan^{-1} x \right)$.
Thus,$y = \cot^{-1} \left( \tan \left( \frac{\pi}{4} + \tan^{-1} x \right) \right)$.
Using $\cot^{-1} \theta = \frac{\pi}{2} - \tan^{-1} \theta$,we have $y = \frac{\pi}{2} - \tan^{-1} \left( \tan \left( \frac{\pi}{4} + \tan^{-1} x \right) \right)$.
$y = \frac{\pi}{2} - \left( \frac{\pi}{4} + \tan^{-1} x \right) = \frac{\pi}{4} - \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \tan^{-1} x \right) = 0 - \frac{1}{1 + x^2} = -\frac{1}{1 + x^2}$.

(B) We know that $1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right)$.
Substituting this into the expression,we get:
$\cos^{-1} \sqrt{\frac{2 \cos^2 (x/2)}{2}} = \cos^{-1} \sqrt{\cos^2 (x/2)} = \cos^{-1} \left( \cos \frac{x}{2} \right)$.
Assuming the principal value branch,$\cos^{-1} (\cos \theta) = \theta$.
Thus,the expression simplifies to $\frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{d}{dx} \left( \frac{x}{2} \right) = \frac{1}{2}$.

(C) Given $y = \tan^{-1}\left( \frac{\sqrt{a} - \sqrt{x}}{1 + \sqrt{a}\sqrt{x}} \right)$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left( \frac{A - B}{1 + AB} \right)$,we can write:
$y = \tan^{-1}(\sqrt{a}) - \tan^{-1}(\sqrt{x})$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}(\sqrt{a})) - \frac{d}{dx}(\tan^{-1}(\sqrt{x}))$.
Since $\sqrt{a}$ is a constant,its derivative is $0$.
Using the chain rule for $\tan^{-1}(\sqrt{x})$:
$\frac{dy}{dx} = 0 - \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x})$.
$\frac{dy}{dx} = - \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}$.
$\frac{dy}{dx} = - \frac{1}{2(1 + x)\sqrt{x}}$.

(A) Given $y = \sec^{-1}\left( \frac{x + 1}{x - 1} \right) + \sin^{-1}\left( \frac{x - 1}{x + 1} \right)$.
We know that $\sec^{-1}(z) = \cos^{-1}\left( \frac{1}{z} \right)$.
Therefore,$\sec^{-1}\left( \frac{x + 1}{x - 1} \right) = \cos^{-1}\left( \frac{x - 1}{x + 1} \right)$.
Substituting this into the expression for $y$,we get:
$y = \cos^{-1}\left( \frac{x - 1}{x + 1} \right) + \sin^{-1}\left( \frac{x - 1}{x + 1} \right)$.
Using the identity $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$,we have:
$y = \frac{\pi}{2}$.
Since $y$ is a constant,its derivative with respect to $x$ is:
$\frac{dy}{dx} = \frac{d}{dx}\left( \frac{\pi}{2} \right) = 0$.

(D) We know that $\tan^{-1}(\cot x) = \tan^{-1}(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$ for $x \in (0, \pi)$.
Similarly,$\cot^{-1}(\tan x) = \cot^{-1}(\cot(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$ for $x \in (0, \pi)$.
Therefore,the expression becomes $\frac{d}{dx}[(\frac{\pi}{2} - x) + (\frac{\pi}{2} - x)]$.
$= \frac{d}{dx}[\pi - 2x]$.
$= 0 - 2 = -2$.

(A) We know the trigonometric identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$ for $\theta \in [-1, 1]$.
Given $y = \sin^{-1}\left(\frac{19}{20}x\right) + \cos^{-1}\left(\frac{19}{20}x\right)$.
By substituting $\theta = \frac{19}{20}x$,we get $y = \frac{\pi}{2}$.
Since $\frac{\pi}{2}$ is a constant,its derivative with respect to $x$ is $0$.
Therefore,$\frac{dy}{dx} = 0$.

(B) Given $y = \tan^{-1}\sqrt{\frac{1 + \cos x}{1 - \cos x}}$.
Using the trigonometric identities $1 + \cos x = 2\cos^2(\frac{x}{2})$ and $1 - \cos x = 2\sin^2(\frac{x}{2})$,we get:
$y = \tan^{-1}\sqrt{\frac{2\cos^2(x/2)}{2\sin^2(x/2)}}$
$y = \tan^{-1}\sqrt{\cot^2(x/2)}$
$y = \tan^{-1}(\cot(x/2))$
Since $\cot(x/2) = \tan(\frac{\pi}{2} - \frac{x}{2})$,we have:
$y = \tan^{-1}(\tan(\frac{\pi}{2} - \frac{x}{2}))$
$y = \frac{\pi}{2} - \frac{x}{2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) - \frac{d}{dx}(\frac{x}{2})$
$\frac{dy}{dx} = 0 - \frac{1}{2} = -\frac{1}{2}$.

(A) Given $y = \cot^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right]$.
We know that $1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2$ and $1 - \sin x = (\cos(x/2) - \sin(x/2))^2$.
Substituting these,we get $y = \cot^{-1} \left[ \frac{(\cos(x/2) + \sin(x/2)) + (\cos(x/2) - \sin(x/2))}{(\cos(x/2) + \sin(x/2)) - (\cos(x/2) - \sin(x/2))} \right]$.
Simplifying the expression inside the bracket: $y = \cot^{-1} \left[ \frac{2\cos(x/2)}{2\sin(x/2)} \right] = \cot^{-1} [\cot(x/2)]$.
Thus,$y = x/2$.
Therefore,$\frac{dy}{dx} = \frac{1}{2}$.

(A) Let $y = \tan^{-1} \left( \frac{a - x}{1 + ax} \right)$.
Using the trigonometric identity $\tan^{-1} \left( \frac{A - B}{1 + AB} \right) = \tan^{-1} A - \tan^{-1} B$,we can rewrite the expression as:
$y = \tan^{-1} a - \tan^{-1} x$.
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\tan^{-1} a) - \frac{d}{dx} (\tan^{-1} x)$.
Since $\tan^{-1} a$ is a constant,its derivative is $0$.
Therefore,$\frac{dy}{dx} = 0 - \frac{1}{1 + x^2} = -\frac{1}{1 + x^2}$.

(C) Given $y = \cos^{-1}\left( \frac{3}{5}\cos x + \frac{4}{5}\sin x \right)$.
Let $\frac{3}{5} = \sin \alpha$ and $\frac{4}{5} = \cos \alpha$,where $\alpha = \sin^{-1}\left( \frac{3}{5} \right)$.
Then $y = \cos^{-1}(\sin \alpha \cos x + \cos \alpha \sin x)$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $y = \cos^{-1}(\sin(x+\alpha))$.
Since $\cos^{-1}(\sin \theta) = \frac{\pi}{2} - \theta$,we have $y = \frac{\pi}{2} - (x+\alpha) = \frac{\pi}{2} - x - \alpha$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2} - x - \alpha) = 0 - 1 - 0 = -1$.

(C) Let $y = \cos^{-1} \left( \frac{x - \frac{1}{x}}{x + \frac{1}{x}} \right) = \cos^{-1} \left( \frac{x^2 - 1}{x^2 + 1} \right)$.
Substitute $x = \cot \theta$,then $x^2 = \cot^2 \theta$.
$y = \cos^{-1} \left( \frac{\cot^2 \theta - 1}{\cot^2 \theta + 1} \right) = \cos^{-1} \left( - \frac{1 - \cot^2 \theta}{1 + \cot^2 \theta} \right) = \cos^{-1} (- \cos 2\theta)$.
Using the identity $\cos^{-1}(-z) = \pi - \cos^{-1}(z)$,we get $y = \pi - \cos^{-1}(\cos 2\theta) = \pi - 2\theta$.
Since $x = \cot \theta$,$\theta = \cot^{-1} x$.
Thus,$y = \pi - 2 \cot^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - 2 \left( - \frac{1}{1 + x^2} \right) = \frac{2}{1 + x^2}$.

(A) Let $y = \cos^{-1} \left( \sqrt{\frac{1+x}{2}} \right)$.
Substitute $x = \cos(2\theta)$,which implies $\theta = \frac{1}{2} \cos^{-1}(x)$.
Then,$\sqrt{\frac{1+x}{2}} = \sqrt{\frac{1+\cos(2\theta)}{2}} = \sqrt{\frac{2\cos^2(\theta)}{2}} = \cos(\theta)$.
Thus,$y = \cos^{-1}(\cos(\theta)) = \theta$.
Substituting back for $\theta$,we get $y = \frac{1}{2} \cos^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{1}{2\sqrt{1-x^2}}$.

(C) Let $x = \cos \theta$. Then $\theta = \cos^{-1} x$.
Using the identities $1 + \cos \theta = 2\cos^2(\frac{\theta}{2})$ and $1 - \cos \theta = 2\sin^2(\frac{\theta}{2})$,we have:
$y = \sin^{-1} \left[ \frac{\sqrt{2}\cos(\frac{\theta}{2}) + \sqrt{2}\sin(\frac{\theta}{2})}{2} \right]$
$y = \sin^{-1} \left[ \frac{1}{\sqrt{2}}\cos(\frac{\theta}{2}) + \frac{1}{\sqrt{2}}\sin(\frac{\theta}{2}) \right]$
Since $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we can write:
$y = \sin^{-1} \left[ \sin(\frac{\pi}{4})\cos(\frac{\theta}{2}) + \cos(\frac{\pi}{4})\sin(\frac{\theta}{2}) \right]$
$y = \sin^{-1} \left[ \sin(\frac{\theta}{2} + \frac{\pi}{4}) \right]$
$y = \frac{\theta}{2} + \frac{\pi}{4}$
Substituting $\theta = \cos^{-1} x$:
$y = \frac{1}{2}\cos^{-1} x + \frac{\pi}{4}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} \right) = -\frac{1}{2\sqrt{1-x^2}}$.

(A) Let ${y_1} = {\tan ^{ - 1}}\left( \frac{{2x}}{{1 - {x^2}}} \right)$ and ${y_2} = {\sin ^{ - 1}}\left( \frac{{2x}}{{1 + {x^2}}} \right)$.
Substitute $x = \tan \theta$,where $\theta = {\tan ^{ - 1}}x$.
Then,${y_1} = {\tan ^{ - 1}}\left( \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} \right) = {\tan ^{ - 1}}(\tan 2\theta) = 2\theta = 2{\tan ^{ - 1}}x$.
Similarly,${y_2} = {\sin ^{ - 1}}\left( \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} \right) = {\sin ^{ - 1}}(\sin 2\theta) = 2\theta = 2{\tan ^{ - 1}}x$.
Now,we need to find the derivative $\frac{{d{y_1}}}{{d{y_2}}}$.
Since ${y_1} = 2{\tan ^{ - 1}}x$ and ${y_2} = 2{\tan ^{ - 1}}x$,we have ${y_1} = {y_2}$.
Therefore,$\frac{{d{y_1}}}{{d{y_2}}} = \frac{d}{{d{y_2}}}({y_2}) = 1$.

(B) We have $y = \tan^{-1}\left( \frac{\log e - \log x^2}{\log e + \log x^2} \right) + \tan^{-1}\left( \frac{3 + 2\log x}{1 - 6\log x} \right)$.
Using the properties of logarithms,$\log e = 1$ and $\log x^2 = 2\log x$,so:
$y = \tan^{-1}\left( \frac{1 - 2\log x}{1 + 2\log x} \right) + \tan^{-1}\left( \frac{3 + 2\log x}{1 - 6\log x} \right)$.
Applying the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left( \frac{A-B}{1+AB} \right)$:
$y = (\tan^{-1} 1 - \tan^{-1}(2\log x)) + (\tan^{-1} 3 + \tan^{-1}(2\log x))$.
Simplifying the expression:
$y = \tan^{-1} 1 + \tan^{-1} 3$.
Since $y$ is a constant,its derivative with respect to $x$ is zero:
$\frac{dy}{dx} = 0$.
Consequently,for any $n \ge 1$,the $n$-th order derivative is:
$\frac{d^n y}{dx^n} = 0$.

(D) Given the equation: $\sin ^{-1}\left(\frac{x}{5}\right) + \csc ^{-1}\left(\frac{5}{4}\right) = \frac{\pi}{2}$
We know that $\csc ^{-1}(z) = \sin ^{-1}\left(\frac{1}{z}\right)$. Therefore,$\csc ^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$.
Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}\right) + \sin ^{-1}\left(\frac{4}{5}\right) = \frac{\pi}{2}$
Rearranging the terms: $\sin ^{-1}\left(\frac{x}{5}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}\right)$
Using the identity $\sin ^{-1}(y) + \cos ^{-1}(y) = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$.
So,$\sin ^{-1}\left(\frac{x}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$.
To solve for $x$,convert $\cos ^{-1}\left(\frac{4}{5}\right)$ to $\sin ^{-1}$. Since $\cos \theta = \frac{4}{5}$,the opposite side is $\sqrt{5^2 - 4^2} = 3$,so $\sin \theta = \frac{3}{5}$.
Thus,$\sin ^{-1}\left(\frac{x}{5}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$.
Comparing the arguments,$\frac{x}{5} = \frac{3}{5}$,which gives $x = 3$.

(D) Given the equation: $\tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} \left( \frac{2x}{1 - x^2} \right)$.
Since $|x| < \frac{1}{\sqrt{3}}$,the condition for the formula $\tan ^{-1} \left( \frac{2x}{1 - x^2} \right) = 2 \tan ^{-1} x$ is satisfied.
Substituting this into the equation,we get:
$\tan ^{-1} y = \tan ^{-1} x + 2 \tan ^{-1} x$
$\tan ^{-1} y = 3 \tan ^{-1} x$
Using the identity $3 \tan ^{-1} x = \tan ^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ for $|x| < \frac{1}{\sqrt{3}}$,we have:
$\tan ^{-1} y = \tan ^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$
Therefore,$y = \frac{3x - x^3}{1 - 3x^2}$.

(B) We are given the expression $\theta = \sin^{-1}x + \cos^{-1}x - \tan^{-1}x$ for $x \ge 0$.
Using the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression simplifies to:
$\theta = \frac{\pi}{2} - \tan^{-1}x$.
Since $x \ge 0$,the range of $\tan^{-1}x$ is $0 \le \tan^{-1}x < \frac{\pi}{2}$.
Multiplying by $-1$,we get $-\frac{\pi}{2} < -\tan^{-1}x \le 0$.
Adding $\frac{\pi}{2}$ to all parts,we get $\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - \tan^{-1}x \le \frac{\pi}{2} - 0$.
Thus,$0 < \theta \le \frac{\pi}{2}$.
Therefore,the correct interval is $0 < \theta \le \frac{\pi}{2}$.

(A) We know that the range of $\sin^{-1} \theta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since the sum of three such values is $\frac{3\pi}{2}$,each term must attain its maximum value.
Therefore,$\sin^{-1} x = \frac{\pi}{2}$,$\sin^{-1} y = \frac{\pi}{2}$,and $\sin^{-1} z = \frac{\pi}{2}$.
This implies $x = \sin(\frac{\pi}{2}) = 1$,$y = 1$,and $z = 1$.
Substituting these values into the expression: $1^{100} + 1^{100} + 1^{100} - \frac{9}{1^{101} + 1^{101} + 1^{101}}$.
$= 1 + 1 + 1 - \frac{9}{1 + 1 + 1} = 3 - \frac{9}{3} = 3 - 3 = 0$.

(D) Let $\cot ^{ - 1}(1/2) = \theta$. Then $\cot \theta = 1/2$.
Since $\cot \theta = 1/2$,we have a right-angled triangle with adjacent side $1$ and opposite side $2$.
The hypotenuse is $\sqrt{1^2 + 2^2} = \sqrt{5}$.
Thus,$\sin \theta = 2/\sqrt{5}$.
Now,the equation is $\tan (\cos ^{ - 1}x) = \sin \theta = 2/\sqrt{5}$.
Let $\cos ^{ - 1}x = \phi$,so $\cos \phi = x$.
Then $\tan \phi = 2/\sqrt{5}$.
In a right-angled triangle with $\tan \phi = 2/\sqrt{5}$,the opposite side is $2$ and the adjacent side is $\sqrt{5}$.
The hypotenuse is $\sqrt{2^2 + (\sqrt{5})^2} = \sqrt{4 + 5} = \sqrt{9} = 3$.
Therefore,$\cos \phi = \text{adjacent} / \text{hypotenuse} = \sqrt{5}/3$.
Hence,$x = \sqrt{5}/3$.

(C) Given that $A = \tan^{-1} 2$ and $B = \tan^{-1} 3$.
We know that in a triangle $ABC$,$A + B + C = \pi$.
Substituting the values,we get $\tan^{-1} 2 + \tan^{-1} 3 + C = \pi$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \left( \frac{x+y}{1-xy} \right)$ for $xy > 1$,we have:
$\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1} \left( \frac{2+3}{1-2 \times 3} \right) = \pi + \tan^{-1} \left( \frac{5}{-5} \right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
Now,$\frac{3\pi}{4} + C = \pi$.
Therefore,$C = \pi - \frac{3\pi}{4} = \frac{\pi}{4}$.

(A) We are given $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$.
Using the identity $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$,we get:
$\cos^{-1} \left[ \frac{x}{a} \cdot \frac{y}{b} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} \right] = \alpha$.
Taking $\cos$ on both sides:
$\frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} = \cos \alpha$.
Rearranging the terms:
$\frac{xy}{ab} - \cos \alpha = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$.
Squaring both sides:
$\left( \frac{xy}{ab} - \cos \alpha \right)^2 = \left( 1 - \frac{x^2}{a^2} \right) \left( 1 - \frac{y^2}{b^2} \right)$.
$\frac{x^2 y^2}{a^2 b^2} - \frac{2xy}{ab} \cos \alpha + \cos^2 \alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2 y^2}{a^2 b^2}$.
Canceling $\frac{x^2 y^2}{a^2 b^2}$ from both sides:
$-\frac{2xy}{ab} \cos \alpha + \cos^2 \alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}$.
Rearranging to isolate the required expression:
$\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = 1 - \cos^2 \alpha = \sin^2 \alpha$.

(B) Given the equation ${x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - x\cos \beta - \sin \beta = 0$.
From Vieta's formulas,we have:
$S_1 = \Sigma {x_1} = \sin 2\beta$
$S_2 = \Sigma {x_1}{x_2} = \cos 2\beta$
$S_3 = \Sigma {x_1}{x_2}{x_3} = \cos \beta$
$S_4 = {x_1}{x_2}{x_3}{x_4} = - \sin \beta$
We know that ${\tan ^{ - 1}}{x_1} + {\tan ^{ - 1}}{x_2} + {\tan ^{ - 1}}{x_3} + {\tan ^{ - 1}}{x_4} = {\tan ^{ - 1}}\left( {\frac{{S_1 - S_3}}{{1 - S_2 + S_4}}} \right)$.
Substituting the values:
$= {\tan ^{ - 1}}\left( {\frac{{\sin 2\beta - \cos \beta }}{{1 - \cos 2\beta - \sin \beta }}} \right)$
$= {\tan ^{ - 1}}\left( {\frac{{2\sin \beta \cos \beta - \cos \beta }}{{1 - (1 - 2\sin^2 \beta) - \sin \beta }}} \right)$
$= {\tan ^{ - 1}}\left( {\frac{{\cos \beta (2\sin \beta - 1)}}{{2\sin^2 \beta - \sin \beta }}} \right)$
$= {\tan ^{ - 1}}\left( {\frac{{\cos \beta (2\sin \beta - 1)}}{{\sin \beta (2\sin \beta - 1)}}} \right)$
$= {\tan ^{ - 1}}(\cot \beta ) = {\tan ^{ - 1}}\left[ {\tan \left( {\frac{\pi }{2} - \beta } \right)} \right] = \frac{\pi }{2} - \beta $.

(B) We know that ${\sin ^{ - 1}}y + {\cos ^{ - 1}}y = \frac{\pi }{2}$ for $|y| \le 1$.
Given the equation ${\sin ^{ - 1}}\left( {x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots} \right) + {\cos ^{ - 1}}\left( {{x^2} - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots} \right) = \frac{\pi }{2}$,it implies that the arguments of the inverse trigonometric functions must be equal.
Let $y = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{4} - \dots$. This is a geometric series with first term $a = x$ and common ratio $r = -x/2$. The sum is $\frac{x}{1 - (-x/2)} = \frac{x}{1 + x/2} = \frac{2x}{2 + x}$.
Similarly,the second argument is $x^2 - \frac{{{x^4}}}{2} + \frac{{{x^6}}}{4} - \dots$,which is a geometric series with $a = x^2$ and $r = -x^2/2$. The sum is $\frac{x^2}{1 + x^2/2} = \frac{2x^2}{2 + x^2}$.
Equating the two: $\frac{2x}{2 + x} = \frac{2x^2}{2 + x^2}$.
Since $x \neq 0$,we can divide by $2x$: $\frac{1}{2 + x} = \frac{x}{2 + x^2}$.
Cross-multiplying gives $2 + x^2 = 2x + x^2$,which simplifies to $2 = 2x$,so $x = 1$.

(D) Given that $\sec^{-1} x = \csc^{-1} y$.
We know that $\sec^{-1} x = \cos^{-1} \frac{1}{x}$ and $\csc^{-1} y = \sin^{-1} \frac{1}{y}$.
Substituting these into the given equation,we get $\cos^{-1} \frac{1}{x} = \sin^{-1} \frac{1}{y}$.
We also know the identity $\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2}$,which implies $\sin^{-1} \frac{1}{y} = \frac{\pi}{2} - \cos^{-1} \frac{1}{y}$.
Substituting this into our equation: $\cos^{-1} \frac{1}{x} = \frac{\pi}{2} - \cos^{-1} \frac{1}{y}$.
Rearranging the terms,we get $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = \frac{\pi}{2}$.

(D) Let $a = \tan \theta$,$b = \tan \phi$,and $x = \tan \psi$.
Substituting these into the given equation:
$\sin ^{ - 1}(\sin 2\theta) - \cos ^{ - 1}(\cos 2\phi) = \tan ^{ - 1}(\tan 2\psi)$
$2\theta - 2\phi = 2\psi$
$\theta - \phi = \psi$
Taking the tangent of both sides:
$\tan(\theta - \phi) = \tan \psi$
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$\frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} = \tan \psi$
Substituting back $a, b, x$:
$\frac{a - b}{1 + ab} = x$

(A) We know that $\sec^{-1}(u) = \cos^{-1}(\frac{1}{u})$ for $|u| \geq 1$.
Given $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}\right)$.
Using the identity $\sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}\right) = \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right)$.
Substituting this into the expression for $y$,we get $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \cos^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right)$.
Since $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$ for $|\theta| \leq 1$,and here $\left|\frac{x^2 - 1}{x^2 + 1}\right| < 1$ for all $x$,we have $y = \frac{\pi}{2}$.
Since $y$ is a constant,the derivative $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

(A) Given the equation: $\tan^{-1}(x+2) + \tan^{-1}(x-2) = \tan^{-1}(\frac{1}{2})$
Using the formula $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}(\frac{A+B}{1-AB})$,we get:
$\tan^{-1}(\frac{(x+2) + (x-2)}{1 - (x+2)(x-2)}) = \tan^{-1}(\frac{1}{2})$
Simplifying the expression inside the $\tan^{-1}$ function:
$\frac{2x}{1 - (x^2 - 4)} = \frac{1}{2}$
$\frac{2x}{1 - x^2 + 4} = \frac{1}{2}$
$\frac{2x}{5 - x^2} = \frac{1}{2}$
Cross-multiplying gives:
$4x = 5 - x^2$
$x^2 + 4x - 5 = 0$
Factoring the quadratic equation:
$(x+5)(x-1) = 0$
So,$x = 1$ or $x = -5$.
Checking the values:
If $x = -5$,then $\tan^{-1}(-3) + \tan^{-1}(-7)$,which is negative,while $\tan^{-1}(\frac{1}{2})$ is positive. Thus,$x = -5$ is rejected.
If $x = 1$,then $\tan^{-1}(3) + \tan^{-1}(-1) = \tan^{-1}(3) - \tan^{-1}(1) = \tan^{-1}(\frac{3-1}{1+3}) = \tan^{-1}(\frac{2}{4}) = \tan^{-1}(\frac{1}{2})$. This is valid.
The only valid value is $x = 1$. The sum of the values is $1$.

(C) We know that $\sin 2 = 2 \sin 1 \cos 1$ and $\cos 2 = \cos^2 1 - \sin^2 1$. Also,$1 = \sin^2 1 + \cos^2 1$.
Substituting these in the expression:
$\tan^{-1} \left( \frac{2 \sin 1 \cos 1 - (\sin^2 1 + \cos^2 1)}{\cos^2 1 - \sin^2 1} \right)$
$= \tan^{-1} \left( \frac{-( \cos^2 1 - 2 \sin 1 \cos 1 + \sin^2 1)}{\cos^2 1 - \sin^2 1} \right)$
$= \tan^{-1} \left( \frac{-(\cos 1 - \sin 1)^2}{(\cos 1 - \sin 1)(\cos 1 + \sin 1)} \right)$
$= \tan^{-1} \left( -\frac{\cos 1 - \sin 1}{\cos 1 + \sin 1} \right)$
$= \tan^{-1} \left( \frac{\sin 1 - \cos 1}{\cos 1 + \sin 1} \right)$
Dividing numerator and denominator by $\cos 1$:
$= \tan^{-1} \left( \frac{\tan 1 - 1}{1 + \tan 1} \right)$
$= \tan^{-1} \left( \tan(1 - \frac{\pi}{4}) \right)$
$= 1 - \frac{\pi}{4}$

(C) The general term of the series is $T_k = \tan^{-1} \frac{1}{1+k(k+1)}$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$,we can write:
$T_k = \tan^{-1} \frac{(k+1)-k}{1+k(k+1)} = \tan^{-1}(k+1) - \tan^{-1}(k)$.
Summing from $k=1$ to $n$:
$S_n = \sum_{k=1}^{n} (\tan^{-1}(k+1) - \tan^{-1}(k))$
$= (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1}(n+1) - \tan^{-1} n)$
$= \tan^{-1}(n+1) - \tan^{-1} 1$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$:
$S_n = \tan^{-1} \frac{(n+1)-1}{1+(n+1)(1)} = \tan^{-1} \frac{n}{n+2}$.
Given $S_n = \tan^{-1} \theta$,we have $\theta = \frac{n}{n+2}$.

(D) We know that $2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2}$.
First,calculate $2 \tan^{-1} \frac{1}{5}$:
$2 \tan^{-1} \frac{1}{5} = \tan^{-1} \frac{2(1/5)}{1-(1/5)^2} = \tan^{-1} \frac{2/5}{24/25} = \tan^{-1} \frac{5}{12}$.
Now,calculate $4 \tan^{-1} \frac{1}{5} = 2(2 \tan^{-1} \frac{1}{5}) = 2 \tan^{-1} \frac{5}{12}$:
$2 \tan^{-1} \frac{5}{12} = \tan^{-1} \frac{2(5/12)}{1-(5/12)^2} = \tan^{-1} \frac{10/12}{1-25/144} = \tan^{-1} \frac{5/6}{119/144} = \tan^{-1} \left( \frac{5}{6} \times \frac{144}{119} \right) = \tan^{-1} \frac{120}{119}$.
Finally,subtract $\tan^{-1} \frac{1}{239}$:
$\tan^{-1} \frac{120}{119} - \tan^{-1} \frac{1}{239} = \tan^{-1} \left( \frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} \times \frac{1}{239}} \right)$,
using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$.
$= \tan^{-1} \left( \frac{28680 - 119}{28441 + 120} \right) = \tan^{-1} \left( \frac{28561}{28561} \right) = \tan^{-1} (1) = \frac{\pi}{4}$.

(C) Let $x = \tan^{-1} \frac{3}{4}$,then $\tan x = \frac{3}{4}$. Using $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$,we get $\cos(2 \tan^{-1} \frac{3}{4}) = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.
Let $y = \cot^{-1} \frac{1}{2}$,then $\cot y = \frac{1}{2}$,so $\tan y = 2$. Using $\sin 2y = \frac{2 \tan y}{1 + \tan^2 y}$,we get $\sin(2 \cot^{-1} \frac{1}{2}) = \frac{2(2)}{1 + 2^2} = \frac{4}{5} = \frac{20}{25}$.
Substituting these values into the expression: $\tan^{-1} [\frac{7}{25} + \frac{20}{25}] = \tan^{-1} [\frac{27}{25}]$.
Since $\frac{27}{25} > 1$,it follows that $\tan^{-1}(\frac{27}{25}) > \tan^{-1}(1)$,which is $\frac{\pi}{4}$.
Thus,the value is greater than $\frac{\pi}{4}$.

(C) Let $x = \cos \theta$. Since $0 < x < 1$,we have $\theta \in (0, \pi/2)$.
Substituting $x = \cos \theta$ into the expression:
$\cot ^{-1}\left( \frac{2\cos ^2 \theta - 1}{2\cos \theta \sqrt{1 - \cos ^2 \theta}} \right)$
$= \cot ^{-1}\left( \frac{\cos 2\theta}{2\cos \theta \sin \theta} \right)$
$= \cot ^{-1}\left( \frac{\cos 2\theta}{\sin 2\theta} \right)$
$= \cot ^{-1}(\cot 2\theta)$
Since $\theta \in (0, \pi/2)$,then $2\theta \in (0, \pi)$.
Thus,$\cot ^{-1}(\cot 2\theta) = 2\theta = 2\cos ^{-1}x$.

(A) We know that $\cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x$.
Substituting this into the given equation:
$(\tan^{-1} x)^2 + (\frac{\pi}{2} - \tan^{-1} x)^2 = \frac{5\pi^2}{8}$.
Let $t = \tan^{-1} x$. Then the equation becomes:
$t^2 + (\frac{\pi}{2} - t)^2 = \frac{5\pi^2}{8}$.
Expanding the square:
$t^2 + \frac{\pi^2}{4} - \pi t + t^2 = \frac{5\pi^2}{8}$.
$2t^2 - \pi t + \frac{\pi^2}{4} - \frac{5\pi^2}{8} = 0$.
$2t^2 - \pi t - \frac{3\pi^2}{8} = 0$.
Multiplying by $8$ to clear the denominator:
$16t^2 - 8\pi t - 3\pi^2 = 0$.
Factoring the quadratic equation:
$(4t - 3\pi)(4t + \pi) = 0$.
So,$t = \frac{3\pi}{4}$ or $t = -\frac{\pi}{4}$.
Since the range of $\tan^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,we must have $\tan^{-1} x = -\frac{\pi}{4}$.
Therefore,$x = \tan(-\frac{\pi}{4}) = -1$.

(D) Let $y = \log(x^2)$. Then $\log(e/x^2) = \log e - \log x^2 = 1 - y$ and $\log(ex^2) = \log e + \log x^2 = 1 + y$.
Statement-$2$: The expression is ${\tan ^{ - 1}}\left[ {\frac{{1 + y}}{{1 - y}}} \right]$. Using the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} \right)$,we have ${\tan ^{ - 1}}1 + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{1 + y}}{{1 - 1 \cdot y}}} \right) = {\tan ^{ - 1}}\left( {\frac{{1 + y}}{{1 - y}}} \right)$. Thus,Statement-$2$ is true.
Statement-$1$: Let $u = \frac{1-y}{1+y}$. The expression is ${\cot ^{ - 1}}(u) + {\cot ^{ - 1}}(1/u)$. We know that ${\cot ^{ - 1}}(u) = {\tan ^{ - 1}}(1/u)$ for $u > 0$. If $u > 0$,then ${\tan ^{ - 1}}(1/u) + {\tan ^{ - 1}}(u) = \frac{\pi}{2}$. If $u < 0$,${\cot ^{ - 1}}(u) = \pi + {\tan ^{ - 1}}(1/u)$,leading to a different result. However,for the identity to hold generally as $\frac{\pi}{2}$,the arguments must be positive. Since Statement-$2$ provides the decomposition,it acts as the basis for the simplification of the terms in Statement-$1$. Thus,Statement-$1$ is true and Statement-$2$ is the correct explanation.

(D) Given $f(n) = \tan^{-1} \left( \frac{e-1}{e^{-n} + e^{n+1}} \right)$.
Multiply the numerator and denominator by $e^n$:
$f(n) = \tan^{-1} \left( \frac{e^n(e-1)}{1 + e^{2n+1}} \right) = \tan^{-1} \left( \frac{e^{n+1} - e^n}{1 + e^{n+1} \cdot e^n} \right)$.
Using the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,we get:
$f(n) = \tan^{-1}(e^{n+1}) - \tan^{-1}(e^n)$.
Now,the sum is a telescoping series:
$S_N = \sum_{n=1}^N f(n) = (\tan^{-1}(e^2) - \tan^{-1}(e)) + (\tan^{-1}(e^3) - \tan^{-1}(e^2)) + \dots + (\tan^{-1}(e^{N+1}) - \tan^{-1}(e^N))$.
$S_N = \tan^{-1}(e^{N+1}) - \tan^{-1}(e)$.
As $N \to \infty$,$e^{N+1} \to \infty$,so $\tan^{-1}(e^{N+1}) \to \frac{\pi}{2}$.
Thus,$\sum_{n=1}^\infty f(n) = \frac{\pi}{2} - \tan^{-1}(e) = \cot^{-1}(e) = \tan^{-1}(\frac{1}{e})$.

(D) The general term of the series is $T_r = \tan ^{-1}\left(\frac{1}{1 + r(r+1)}\right)$.
Using the identity $\tan ^{-1}(x) - \tan ^{-1}(y) = \tan ^{-1}\left(\frac{x-y}{1+xy}\right)$,we can write $T_r = \tan ^{-1}(r+1) - \tan ^{-1}(r)$.
Summing from $r=1$ to $r=5$:
$S = \sum_{r=1}^{5} (\tan ^{-1}(r+1) - \tan ^{-1}(r)) = (\tan ^{-1} 2 - \tan ^{-1} 1) + (\tan ^{-1} 3 - \tan ^{-1} 2) + \dots + (\tan ^{-1} 6 - \tan ^{-1} 5)$.
This is a telescoping series,so $S = \tan ^{-1} 6 - \tan ^{-1} 1$.
$S = \tan ^{-1}\left(\frac{6-1}{1+6 \times 1}\right) = \tan ^{-1}\left(\frac{5}{7}\right)$.
Thus,$p = 5$ and $q = 7$. Since $p$ and $q$ are relatively prime,$p + q = 5 + 7 = 12$.

(A) Let $g(x) = |\sin^{-1}|\sin x|| - |\cos^{-1}|\cos x||$.
We know that for any $x \in R$,$|\sin x| \in [0, 1]$ and $|\cos x| \in [0, 1]$.
Thus,$\sin^{-1}|\sin x| \in [0, \frac{\pi}{2}]$ and $\cos^{-1}|\cos x| \in [0, \frac{\pi}{2}]$.
Since both terms are non-negative,we can write $g(x) = \sin^{-1}|\sin x| - \cos^{-1}|\cos x|$.
Using the identity $\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2}$,we have $\sin^{-1}|\sin x| = \frac{\pi}{2} - \cos^{-1}|\sin x|$.
Also,$\cos^{-1}|\cos x| = \sin^{-1} \sqrt{1 - \cos^2 x} = \sin^{-1}|\sin x|$.
Therefore,$g(x) = \sin^{-1}|\sin x| - \sin^{-1}|\sin x| = 0$.
Since $g(x) = 0$ for all $x \in R$,the function $f(x) = \sqrt{0} = 0$.
Thus,the range of the function is $\{0\}$.