Properties of ITF Questions in English

(A) Given $y = \sin^{-1} \left( \frac{5}{13}x + \frac{12}{13}\sqrt{1-x^2} \right)$.
Let $x = \sin \theta$,then $\sqrt{1-x^2} = \cos \theta$.
Let $\cos \alpha = \frac{5}{13}$,then $\sin \alpha = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$.
Substituting these into the equation:
$y = \sin^{-1} (\sin \theta \cos \alpha + \cos \theta \sin \alpha)$
$y = \sin^{-1} (\sin(\theta + \alpha))$
$y = \theta + \alpha$
Since $\theta = \sin^{-1} x$ and $\alpha = \cos^{-1} (\frac{5}{13})$ (a constant),
$y = \sin^{-1} x + \cos^{-1} (\frac{5}{13})$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} (\sin^{-1} x) + \frac{d}{dx} (\cos^{-1} (\frac{5}{13}))$
$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} + 0 = \frac{1}{\sqrt{1-x^2}}$.

(A) We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$ for $|z| \geq 1$.
Given $y = \operatorname{cosec}^{-1}\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$.
Using the identity,we can rewrite the first term as $\sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$.
Thus,$y = \sin^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right) + \cos^{-1}\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)$.
Using the inverse trigonometric identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$,we get $y = \frac{\pi}{2}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

(A) We use the properties of inverse trigonometric functions:
$1$. $\sin ^{-1}(-x) = -\sin ^{-1}(x)$
$2$. $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$
$3$. $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$
$4$. $\tan ^{-1}(x)$ is standard.
Evaluating each term:
$\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$
$\cos ^{-1}\left(-\frac{1}{2}\right) = \pi - \cos ^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \pi - \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\tan ^{-1}(\sqrt{3}) = \frac{\pi}{3}$
Substituting these values into the expression:
$-\frac{\pi}{4} + \frac{2\pi}{3} - \frac{2\pi}{3} + \frac{\pi}{3} = -\frac{\pi}{4} + \frac{\pi}{3} = \frac{-\pi + 4\pi}{12} = \frac{\pi}{12}$
Thus,the correct option is $A$.

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

(A) Given that $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$.
Since the range of $\cos ^{-1} \theta$ is $[0, \pi]$,the maximum value of each term is $\pi$.
For the sum of three terms to be $3 \pi$,each term must be equal to its maximum value:
$\cos ^{-1} x = \pi$,$\cos ^{-1} y = \pi$,and $\cos ^{-1} z = \pi$.
This implies $x = \cos \pi = -1$,$y = \cos \pi = -1$,and $z = \cos \pi = -1$.
Now,substitute $x = -1$ into the expression $x^{2025}+x^{2026}+x^{2027}$:
$(-1)^{2025} + (-1)^{2026} + (-1)^{2027} = -1 + 1 - 1 = -1$.

(D) Given the equation: $\tan ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$
Since $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$,we can write:
$\tan ^{-1} \sqrt{x^2+x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x^2+x+1} = \cos ^{-1} \sqrt{x^2+x+1}$
Let $\tan ^{-1} \sqrt{x^2+x} = \theta$. Then $\tan \theta = \sqrt{x^2+x}$.
This implies $\cos \theta = \frac{1}{\sqrt{1 + (x^2+x)}}$.
Thus,$\cos ^{-1} \left( \frac{1}{\sqrt{x^2+x+1}} \right) = \cos ^{-1} \sqrt{x^2+x+1}$.
Comparing the arguments: $\frac{1}{\sqrt{x^2+x+1}} = \sqrt{x^2+x+1}$
$1 = x^2+x+1$
$x^2+x = 0$
$x(x+1) = 0$
So,$x = 0$ or $x = -1$.
Checking the domain of $\sin ^{-1} \sqrt{x^2+x+1}$,we require $0 \le x^2+x+1 \le 1$,which implies $x^2+x \le 0$.
For $x=0$,$x^2+x=0$ (valid). For $x=-1$,$x^2+x=0$ (valid).
However,$\tan ^{-1} \sqrt{x^2+x}$ requires $x^2+x \ge 0$.
Thus,$x^2+x=0$ is the only solution,giving $x=0$ or $x=-1$.

(A) Let $x = \tan \theta$ and $y = \tan \phi$.
Substituting these into the expression:
$\frac{1}{2} \sin ^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = \frac{1}{2} \sin ^{-1} (\sin 2 \theta) = \frac{1}{2} (2 \theta) = \theta$.
Similarly,$\frac{1}{2} \cos ^{-1} \left( \frac{1 - \tan^2 \phi}{1 + \tan^2 \phi} \right) = \frac{1}{2} \cos ^{-1} (\cos 2 \phi) = \frac{1}{2} (2 \phi) = \phi$.
Now,the expression becomes $\tan (\theta + \phi)$.
Using the trigonometric identity $\tan (\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}$,we get:
$\frac{x + y}{1 - xy}$.

(B) Let $\alpha = \cos ^{-1}\left(\frac{4}{5}\right)$ and $\beta = \cos ^{-1}\left(\frac{12}{13}\right)$.
Then $\cos \alpha = \frac{4}{5}$ and $\cos \beta = \frac{12}{13}$.
Using the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$,we get $\sin \alpha = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$ and $\sin \beta = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13}$.
Using the formula $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:
$\cos(\alpha + \beta) = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right)$
$= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$.
Therefore,$\alpha + \beta = \cos ^{-1}\left(\frac{33}{65}\right)$.

(C) Given the equation: $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$:
$\tan ^{-1} \left[ \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left( \frac{x-1}{x-2} \right) \left( \frac{x+1}{x+2} \right)} \right] = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) - (x-1)(x+1)} = \tan \frac{\pi}{4} = 1$
Expanding the terms:
$\frac{(x^2+x-2) + (x^2-x-2)}{(x^2-4) - (x^2-1)} = 1$
$\frac{2x^2 - 4}{-4 + 1} = 1$
$\frac{2x^2 - 4}{-3} = 1$
$2x^2 - 4 = -3$
$2x^2 = 1$
$x^2 = \frac{1}{2}$
$x = \pm \frac{1}{\sqrt{2}}$

(A) Let $\sin ^{-1}\left(\frac{3}{5}\right) = x$ and $\cos ^{-1}\left(\frac{12}{13}\right) = y$.
Then $\sin x = \frac{3}{5} \implies \cos x = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.
And $\cos y = \frac{12}{13} \implies \sin y = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$.
We are given $x + y = \sin ^{-1} \alpha$,which means $\sin(x + y) = \alpha$.
Using the identity $\sin(x + y) = \sin x \cos y + \cos x \sin y$:
$\alpha = (\frac{3}{5}) \times (\frac{12}{13}) + (\frac{4}{5}) \times (\frac{5}{13})$
$\alpha = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}$.

(C) Given equation: $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$.
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the equation:
$4 \sin ^{-1} x + 6(\frac{\pi}{2} - \sin ^{-1} x) = 3 \pi$
$4 \sin ^{-1} x + 3 \pi - 6 \sin ^{-1} x = 3 \pi$
$-2 \sin ^{-1} x + 3 \pi = 3 \pi$
$-2 \sin ^{-1} x = 0$
$\sin ^{-1} x = 0$
$x = \sin(0) = 0$.

(B) Let $L = \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} + \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8}$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$:
First,group the terms: $L = \left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} \right) + \left( \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8} \right)$.
Calculate the first part: $\tan ^{-1} \left( \frac{1/3 + 1/5}{1 - (1/3)(1/5)} \right) = \tan ^{-1} \left( \frac{8/15}{14/15} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$.
Calculate the second part: $\tan ^{-1} \left( \frac{1/7 + 1/8}{1 - (1/7)(1/8)} \right) = \tan ^{-1} \left( \frac{15/56}{55/56} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$.
Now combine them: $L = \tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{4/7 + 3/11}{1 - (4/7)(3/11)} \right)$.
$L = \tan ^{-1} \left( \frac{44/77 + 21/77}{1 - 12/77} \right) = \tan ^{-1} \left( \frac{65/77}{65/77} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$.

(C) Given the equation: $\tan ^{-1} 2x + \tan ^{-1} 3x = \frac{\pi}{4}$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} \right) = \frac{\pi}{4}$
$\tan ^{-1} \left( \frac{5x}{1 - 6x^2} \right) = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{5x}{1 - 6x^2} = \tan \left( \frac{\pi}{4} \right) = 1$
$5x = 1 - 6x^2$
$6x^2 + 5x - 1 = 0$
Factoring the quadratic equation:
$6x^2 + 6x - x - 1 = 0$
$6x(x + 1) - 1(x + 1) = 0$
$(6x - 1)(x + 1) = 0$
This gives $x = \frac{1}{6}$ or $x = -1$.
Since $\tan ^{-1} 2x + \tan ^{-1} 3x = \frac{\pi}{4} > 0$,$x$ must be positive. Therefore,$x = -1$ is rejected.
Thus,$x = \frac{1}{6}$.

(A) Let $y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$.
Substitute $x = \tan \theta$,so $\theta = \tan^{-1} x$.
Then $y = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right) = \tan^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$.
$y = \tan^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan^{-1}(\tan(\theta/2))$.
$y = \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)}$.

(C) We are given the equation $4 \sin ^{-1} x + \cos ^{-1} x = \pi$.
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the given equation:
$4 \sin ^{-1} x + (\frac{\pi}{2} - \sin ^{-1} x) = \pi$
$3 \sin ^{-1} x + \frac{\pi}{2} = \pi$
$3 \sin ^{-1} x = \frac{\pi}{2}$
$\sin ^{-1} x = \frac{\pi}{6}$
Taking the sine of both sides:
$x = \sin(\frac{\pi}{6}) = \frac{1}{2}$.
Thus,the correct option is $C$.

(C) Let $x = \cos \theta$. Then $\theta = \cos^{-1} x$.
Substituting $x = \cos \theta$ in the expression,we get $\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \tan(\theta/2)$.
So,the expression becomes $y = \sin^2 \left( \cot^{-1} (\tan(\theta/2)) \right)$.
Since $\tan(\theta/2) = \cot(\frac{\pi}{2} - \frac{\theta}{2})$,we have $y = \sin^2 \left( \cot^{-1} \left( \cot \left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) \right)$.
This simplifies to $y = \sin^2 \left( \frac{\pi}{2} - \frac{\theta}{2} \right) = \cos^2(\theta/2)$.
Using the identity $\cos^2(\theta/2) = \frac{1+\cos \theta}{2}$,we get $y = \frac{1+x}{2} = \frac{1}{2} + \frac{x}{2}$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} + \frac{x}{2} \right) = 0 + \frac{1}{2} = \frac{1}{2}$.

(B) Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
Substituting this into the expression:
$\tan^{-1}\left(\frac{1-\tan^2 \theta}{2 \tan \theta}\right) = \tan^{-1}\left(\frac{1}{\tan 2\theta}\right) = \tan^{-1}(\cot 2\theta) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta$.
Next,$\cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$.
Adding these two parts: $(\frac{\pi}{2} - 2\theta) + 2\theta = \frac{\pi}{2}$.
Finally,$\sin\left(\frac{\pi}{2}\right) = 1$.

(C) Given that $\cot ^{-1}\left(\frac{1}{x}\right)-\cot ^{-1}\left(\frac{1}{y}\right)=0$,which implies $\cot ^{-1}\left(\frac{1}{x}\right)=\cot ^{-1}\left(\frac{1}{y}\right)$.
This means $\frac{1}{x}=\frac{1}{y}$,so $x=y$.
Substitute $x=y$ into the first equation: $\sin ^{-1} x+\sin ^{-1} x=\frac{\pi}{3}$.
$2 \sin ^{-1} x=\frac{\pi}{3} \implies \sin ^{-1} x=\frac{\pi}{6}$.
Therefore,$x=\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}$.
Since $x=y$,we have $y=\frac{1}{2}$.
Now,calculate $2 x^2+y^2-x y$:
$2 \left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = 2 \left(\frac{1}{4}\right)+\frac{1}{4}-\frac{1}{4} = \frac{1}{2}+0 = \frac{1}{2}$.

(C) Let $\theta_1 = \sec ^{-1} 4$,then $\sec \theta_1 = 4$.
Using the identity $\tan ^2 \theta = \sec ^2 \theta - 1$,we get $\tan ^2(\sec ^{-1} 4) = \sec ^2(\sec ^{-1} 4) - 1 = 4^2 - 1 = 16 - 1 = 15$.
Let $\theta_2 = \operatorname{cosec}^{-1} 3$,then $\operatorname{cosec} \theta_2 = 3$.
Using the identity $\cot ^2 \theta = \operatorname{cosec}^2 \theta - 1$,we get $\cot ^2(\operatorname{cosec}^{-1} 3) = \operatorname{cosec}^2(\operatorname{cosec}^{-1} 3) - 1 = 3^2 - 1 = 9 - 1 = 8$.
Adding these values,we get $15 + 8 = 23$.

(A) We are given the expression $\cos(2 \cos^{-1} x + \sin^{-1} x)$.
We can rewrite this as $\cos(\cos^{-1} x + (\cos^{-1} x + \sin^{-1} x))$.
Using the identity $\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression becomes $\cos(\cos^{-1} x + \frac{\pi}{2})$.
Using the trigonometric identity $\cos(\theta + \frac{\pi}{2}) = -\sin \theta$,we get $-\sin(\cos^{-1} x)$.
Since $\sin(\cos^{-1} x) = \sqrt{1 - x^2}$,the expression simplifies to $-\sqrt{1 - x^2}$.
Given $x = \frac{1}{5}$,we substitute this value into the simplified expression:
$-\sqrt{1 - (\frac{1}{5})^2} = -\sqrt{1 - \frac{1}{25}} = -\sqrt{\frac{24}{25}}$.
Thus,the correct option is $A$.

(D) Given the inequality $(\cos ^{-1} x)^2 - (\sin ^{-1} x)^2 > 0$.
Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we get:
$(\cos ^{-1} x - \sin ^{-1} x)(\cos ^{-1} x + \sin ^{-1} x) > 0$.
We know that $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Substituting this,we have:
$(\cos ^{-1} x - \sin ^{-1} x) \cdot \frac{\pi}{2} > 0$.
Since $\frac{\pi}{2} > 0$,we can divide by it:
$\cos ^{-1} x - \sin ^{-1} x > 0$
$\cos ^{-1} x > \sin ^{-1} x$.
Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we substitute:
$\frac{\pi}{2} - \sin ^{-1} x > \sin ^{-1} x$
$\frac{\pi}{2} > 2 \sin ^{-1} x$
$\sin ^{-1} x < \frac{\pi}{4}$
Since $\sin \theta$ is an increasing function,we take the sine of both sides:
$x < \sin(\frac{\pi}{4})$
$x < \frac{1}{\sqrt{2}}$.
Also,the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$.
Therefore,the solution is $-1 \leqslant x < \frac{1}{\sqrt{2}}$.

(B) We need to find the principal value of $\cos ^{-1}\left[\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right]$.
First,rewrite the expression inside the bracket:
$\frac{1}{\sqrt{2}} \cos \frac{9 \pi}{10} - \frac{1}{\sqrt{2}} \sin \frac{9 \pi}{10}$
Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have:
$\cos \frac{\pi}{4} \cos \frac{9 \pi}{10} - \sin \frac{\pi}{4} \sin \frac{9 \pi}{10}$
Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get:
$\cos \left(\frac{\pi}{4} + \frac{9 \pi}{10}\right) = \cos \left(\frac{5 \pi + 18 \pi}{20}\right) = \cos \left(\frac{23 \pi}{20}\right)$
Since $\cos \theta = \cos(2 \pi - \theta)$,we have $\cos \left(\frac{23 \pi}{20}\right) = \cos \left(2 \pi - \frac{23 \pi}{20}\right) = \cos \left(\frac{17 \pi}{20}\right)$.
Thus,$\cos ^{-1}\left[\cos \frac{17 \pi}{20}\right] = \frac{17 \pi}{20}$,which lies in the range $[0, \pi]$.
Therefore,the correct option is $B$.

(B) We know that $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$.
Let $u = \tan ^{-1} x$. Then $\cot ^{-1} x = \frac{\pi}{2} - u$.
The given equation becomes $u^2 + (\frac{\pi}{2} - u)^2 = \frac{5 \pi^2}{8}$.
Expanding the equation: $u^2 + \frac{\pi^2}{4} - \pi u + u^2 = \frac{5 \pi^2}{8}$.
$2u^2 - \pi u + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.
$2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.
Multiply by $8$: $16u^2 - 8\pi u - 3\pi^2 = 0$.
Factoring the quadratic: $(4u - 3\pi)(4u + \pi) = 0$.
So,$u = \frac{3\pi}{4}$ or $u = -\frac{\pi}{4}$.
Since the range of $\tan ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,we must have $u = -\frac{\pi}{4}$.
Thus,$\tan ^{-1} x = -\frac{\pi}{4}$,which implies $x = \tan(-\frac{\pi}{4}) = -1$.
Therefore,$x^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2$.

(A) Given the equation $\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}$.
Since the range of $\sin ^{-1}(y)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,for the sum to be $-\frac{\pi}{2}$,both terms must be negative or zero.
Let $4x = \sin(\alpha)$ and $4\sqrt{3}x = \sin(\beta)$,where $\alpha, \beta \in [-\frac{\pi}{2}, 0]$.
Then $\alpha + \beta = -\frac{\pi}{2}$,which implies $\alpha = -\frac{\pi}{2} - \beta$.
Taking sine on both sides: $\sin(\alpha) = \sin(-\frac{\pi}{2} - \beta) = -\cos(\beta)$.
Since $\sin^2(\beta) + \cos^2(\beta) = 1$,we have $\cos(\beta) = \sqrt{1 - \sin^2(\beta)}$.
Substituting the values: $4x = -\sqrt{1 - (4\sqrt{3}x)^2}$.
Squaring both sides: $16x^2 = 1 - 48x^2$.
$64x^2 = 1 \implies x^2 = \frac{1}{64}$.
Since $x$ must be negative for the sum to be negative,$x = -\frac{1}{8}$.
The absolute value of $x$ is $|x| = \frac{1}{8}$.

(D) Given the equation: $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$.
We know the identity $\cot ^{-1}(y) = \frac{\pi}{2} - \tan ^{-1}(y)$.
Substituting this into the equation: $(\frac{\pi}{2} - \tan ^{-1}(\sqrt{\cos \alpha})) - \tan ^{-1}(\sqrt{\cos \alpha}) = x$.
This simplifies to: $\frac{\pi}{2} - 2 \tan ^{-1}(\sqrt{\cos \alpha}) = x$.
Rearranging gives: $2 \tan ^{-1}(\sqrt{\cos \alpha}) = \frac{\pi}{2} - x$.
Taking the tangent of both sides: $\tan(2 \tan ^{-1}(\sqrt{\cos \alpha})) = \tan(\frac{\pi}{2} - x) = \cot x$.
Using the formula $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,where $\theta = \tan ^{-1}(\sqrt{\cos \alpha})$:
$\cot x = \frac{2 \sqrt{\cos \alpha}}{1 - \cos \alpha}$.
Using the half-angle identity $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$:
$\cot x = \frac{2 \sqrt{\cos \alpha}}{2 \sin^2 \frac{\alpha}{2}} = \frac{\sqrt{\cos \alpha}}{\sin^2 \frac{\alpha}{2}}$.
However,checking the standard identity $2 \tan ^{-1}(y) = \cos ^{-1}(\frac{1-y^2}{1+y^2})$,we have $x = \frac{\pi}{2} - \cos ^{-1}(\frac{1-\cos \alpha}{1+\cos \alpha}) = \sin ^{-1}(\frac{1-\cos \alpha}{1+\cos \alpha})$.
Thus,$\sin x = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2 \sin^2 \frac{\alpha}{2}}{2 \cos^2 \frac{\alpha}{2}} = \tan^2 \frac{\alpha}{2}$.

(D) Let $\cot^{-1} x = \theta$. Then $x = \cot \theta$.
Since $0 < x < 1$,$\theta$ lies in the interval $(\frac{\pi}{4}, \frac{\pi}{2})$.
Using the identity $\csc^2 \theta = 1 + \cot^2 \theta$,we have $\csc \theta = \sqrt{1 + x^2}$,so $\sin \theta = \frac{1}{\sqrt{1 + x^2}}$.
Also,$\cos \theta = \cot \theta \sin \theta = \frac{x}{\sqrt{1 + x^2}}$.
Now,substitute these into the expression:
$E = \sqrt{1 + x^2} [\{x \cos \theta + \sin \theta\} ^2 - 1]^{\frac{1}{2}}$
$E = \sqrt{1 + x^2} [\{x (\frac{x}{\sqrt{1 + x^2}}) + \frac{1}{\sqrt{1 + x^2}}\} ^2 - 1]^{\frac{1}{2}}$
$E = \sqrt{1 + x^2} [(\frac{x^2 + 1}{\sqrt{1 + x^2}})^2 - 1]^{\frac{1}{2}}$
$E = \sqrt{1 + x^2} [(\sqrt{1 + x^2})^2 - 1]^{\frac{1}{2}}$
$E = \sqrt{1 + x^2} (1 + x^2 - 1)^{\frac{1}{2}}$
$E = \sqrt{1 + x^2} (x^2)^{\frac{1}{2}}$
Since $x > 0$,$(x^2)^{\frac{1}{2}} = x$.
Therefore,$E = x \sqrt{1 + x^2}$.

(D) Let $S = \sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}$.
First,convert $\sin ^{-1} \frac{4}{5}$ and $\sin ^{-1} \frac{5}{13}$ to $\tan ^{-1}$ form:
$\sin ^{-1} \frac{4}{5} = \tan ^{-1} \frac{4}{3}$ and $\sin ^{-1} \frac{5}{13} = \tan ^{-1} \frac{5}{12}$.
Now,calculate the sum of the first two terms:
$\tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{5}{12} = \tan ^{-1} \left(\frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} \times \frac{5}{12}}\right) = \tan ^{-1} \left(\frac{\frac{16+5}{12}}{\frac{36-20}{36}}\right) = \tan ^{-1} \left(\frac{21}{12} \times \frac{36}{16}\right) = \tan ^{-1} \left(\frac{63}{16}\right)$.
Since $\tan ^{-1} \frac{63}{16} = \cot ^{-1} \frac{16}{63}$,we can convert this to $\sin ^{-1}$ form as $\sin ^{-1} \frac{63}{65}$.
Alternatively,note that $\tan ^{-1} \frac{63}{16} = \cos ^{-1} \frac{16}{65}$.
Thus,$S = \cos ^{-1} \frac{16}{65} + \sin ^{-1} \frac{16}{65}$.
Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get $S = \frac{\pi}{2}$.
Finally,the expression is $2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$.

(D) Given equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$
We know that $\operatorname{cosec}^{-1}(y) = \sin ^{-1}\left(\frac{1}{y}\right)$.
So,$\operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$.
Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)$
Using the identity $\sin ^{-1}(z)+\cos ^{-1}(z)=\frac{\pi}{2}$,we get $\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$.
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\cos ^{-1}\left(\frac{4}{5}\right)$
Now,convert $\cos ^{-1}\left(\frac{4}{5}\right)$ to $\sin ^{-1}$ form: $\cos ^{-1}\left(\frac{4}{5}\right) = \sin ^{-1}\left(\sqrt{1-\left(\frac{4}{5}\right)^2}\right) = \sin ^{-1}\left(\sqrt{1-\frac{16}{25}}\right) = \sin ^{-1}\left(\sqrt{\frac{9}{25}}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$.
$\Rightarrow \sin ^{-1}\left(\frac{x}{5}\right)=\sin ^{-1}\left(\frac{3}{5}\right)$
Comparing both sides,we get $\frac{x}{5} = \frac{3}{5}$,which implies $x = 3$.

(B) Given the equation: $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$
We know that $\tan ^{-1}(A) + \cot ^{-1}(A) = \frac{\pi}{2}$,so $\frac{\pi}{2} - \tan ^{-1}(1-x) = \cot ^{-1}(1-x)$.
Substituting this into the equation: $\tan ^{-1}(1+x) = \cot ^{-1}(1-x)$.
Using the identity $\cot ^{-1}(y) = \tan ^{-1}(\frac{1}{y})$,we get: $\tan ^{-1}(1+x) = \tan ^{-1}(\frac{1}{1-x})$.
Equating the arguments: $1+x = \frac{1}{1-x}$.
This simplifies to: $(1+x)(1-x) = 1$.
$1 - x^2 = 1$.
$-x^2 = 0$,which implies $x = 0$.

(B) Given $x = \operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$.
Let $\sin^{-1} a = \theta$,so $\sin \theta = a$. Then $\sec \theta = \frac{1}{\sqrt{1-a^2}}$.
Now,$x = \operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\frac{1}{\sqrt{1-a^2}}))))$.
Let $\cot^{-1}(\frac{1}{\sqrt{1-a^2}}) = \phi$,so $\cot \phi = \frac{1}{\sqrt{1-a^2}}$.
Then $\cos \phi = \frac{\cot \phi}{\sqrt{1+\cot^2 \phi}} = \frac{1/\sqrt{1-a^2}}{\sqrt{1 + 1/(1-a^2)}} = \frac{1}{\sqrt{1-a^2+1}} = \frac{1}{\sqrt{2-a^2}}$.
So,${x = \operatorname{cosec}(\tan ^{-1}(\frac{1}{\sqrt{2-a^2}}}))$.
Let $\tan^{-1}(\frac{1}{\sqrt{2-a^2}}) = \psi$,so $\tan \psi = \frac{1}{\sqrt{2-a^2}}$.
Then $\operatorname{cosec} \psi = \sqrt{1+\cot^2 \psi} = \sqrt{1+(2-a^2)} = \sqrt{3-a^2}$.
Thus,$x = \sqrt{3-a^2}$,which implies $x^2 = 3-a^2$,or $x^2+a^2=3$.

(D) Given the equation: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
Since $\sin \frac{\pi}{2} = 1$,we have:
$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$
Using the identity $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$,we can rewrite the equation as:
$\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} \frac{1}{5}$
Since $\frac{\pi}{2} - \sin ^{-1} \theta = \cos ^{-1} \theta$,we get:
$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$
Therefore,$x = \frac{1}{5}$.

(D) Let $x = \tan \theta$. Since $x \in (0, 1)$,we have $\theta \in (0, \frac{\pi}{4})$.
First,consider $A = 2 \tan^{-1}\left(\frac{1+x}{1-x}\right)$.
Using the formula $\tan^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) = \tan^{-1}(\tan(\frac{\pi}{4} + \theta)) = \frac{\pi}{4} + \theta$,we get $A = 2(\frac{\pi}{4} + \theta) = \frac{\pi}{2} + 2\theta$.
Next,consider $B = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Using the formula $\cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$.
Finally,$A - B = (\frac{\pi}{2} + 2\theta) - 2\theta = \frac{\pi}{2}$.

(A) We know the principal values of the inverse trigonometric functions:
$\operatorname{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$
$\cos^{-1}\left(\frac{-1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$
Substituting these values into the expression:
$\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}$
Finding a common denominator $(12)$:
$\frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$

(C) We are given the equation: $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$.
Using the formula for the sum of three inverse tangent functions: $\tan ^{-1} x + \tan ^{-1} y + \tan ^{-1} z = \tan ^{-1} \left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right)$.
Applying this to the given equation: $\tan ^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} \right) = \pi$.
Taking the tangent of both sides: $\frac{a+b+c-abc}{1-ab-bc-ca} = \tan \pi$.
Since $\tan \pi = 0$,we have: $\frac{a+b+c-abc}{1-ab-bc-ca} = 0$.
This implies that the numerator must be zero: $a+b+c-abc = 0$.
Therefore,$a+b+c = abc$.

(B) The principal value range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given that $\sin ^{-1} x + \sin ^{-1} y + \sin ^{-1} z = \frac{3 \pi}{2}$.
Since the maximum value of each $\sin ^{-1}$ term is $\frac{\pi}{2}$,the sum can only be $\frac{3 \pi}{2}$ if each term is individually equal to 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 $x = \sin(\frac{\pi}{2}) = 1$,$y = \sin(\frac{\pi}{2}) = 1$,and $z = \sin(\frac{\pi}{2}) = 1$.
Substituting these values into the expression $x^{100} + y^{100} + z^{100}$:
$1^{100} + 1^{100} + 1^{100} = 1 + 1 + 1 = 3$.

(A) Given the equation: $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$
Using the formula $\tan ^{-1} A+\tan ^{-1} B=\tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$\tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{5 x}{1-6 x^2}\right)=\frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{5 x}{1-6 x^2}=\tan\left(\frac{\pi}{4}\right)=1$
$5 x=1-6 x^2$
$6 x^2+5 x-1=0$
Factoring the quadratic equation:
$(6 x-1)(x+1)=0$
So,$x=\frac{1}{6}$ or $x=-1$.
Checking the values: If $x=-1$,$\tan ^{-1}(-2)+\tan ^{-1}(-3)$ is negative,which cannot be $\frac{\pi}{4}$.
Thus,$x=\frac{1}{6}$ is the only valid solution.

(B) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$.
Using the formula $\cos ^{-1} A - \cos ^{-1} B = \cos ^{-1} (AB + \sqrt{1-A^2} \sqrt{1-B^2})$,we get:
$\cos ^{-1} \left( x \cdot \frac{y}{2} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}} \right) = \alpha$.
Taking cosine on both sides:
$\frac{xy}{2} + \frac{\sqrt{(1-x^2)(4-y^2)}}{2} = \cos \alpha$.
$\sqrt{(1-x^2)(4-y^2)} = 2 \cos \alpha - xy$.
Squaring both sides:
$(1-x^2)(4-y^2) = (2 \cos \alpha - xy)^2$.
$4 - y^2 - 4x^2 + x^2y^2 = 4 \cos ^2 \alpha - 4xy \cos \alpha + x^2y^2$.
Subtracting $x^2y^2$ from both sides:
$4 - y^2 - 4x^2 = 4 \cos ^2 \alpha - 4xy \cos \alpha$.
Rearranging the terms:
$4x^2 - 4xy \cos \alpha + y^2 = 4 - 4 \cos ^2 \alpha$.
Since $1 - \cos ^2 \alpha = \sin ^2 \alpha$,we have:
$4x^2 - 4xy \cos \alpha + y^2 = 4 \sin ^2 \alpha$.

(D) We need to evaluate $\sin ^{-1}[\sin (-600^{\circ})] + \cot ^{-1}(-\sqrt{3})$.
First,simplify $\sin (-600^{\circ})$:
$\sin (-600^{\circ}) = -\sin (600^{\circ}) = -\sin (360^{\circ} + 240^{\circ}) = -\sin (240^{\circ}) = -\sin (180^{\circ} + 60^{\circ}) = -(-\sin 60^{\circ}) = \sin 60^{\circ}$.
Thus,$\sin ^{-1}[\sin (-600^{\circ})] = \sin ^{-1}(\sin 60^{\circ}) = 60^{\circ} = \frac{\pi}{3}$.
Next,evaluate $\cot ^{-1}(-\sqrt{3})$:
Since $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$,we have $\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Adding the two results:
$\frac{\pi}{3} + \frac{5\pi}{6} = \frac{2\pi + 5\pi}{6} = \frac{7\pi}{6}$.

(B) Given the equation: $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$.
Using the formula $2 \tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,we get:
$\tan^{-1}\left(\frac{2 \cos x}{1-\cos^2 x}\right) = \tan^{-1}\left(\frac{2}{\sin x}\right)$.
Since $\tan^{-1}$ is a one-to-one function,we equate the arguments:
$\frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x}$.
Assuming $\sin x \neq 0$,we can divide both sides by $\frac{2}{\sin x}$:
$\frac{\cos x}{\sin x} = 1$,which implies $\cot x = 1$.
Therefore,$x = \frac{\pi}{4}$.

(A) Let $\alpha = \cos^{-1} \frac{1}{5\sqrt{2}}$. Then $\cos \alpha = \frac{1}{5\sqrt{2}}$.
Since $\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{1}{50} = \frac{49}{50}$,we have $\sin \alpha = \frac{7}{5\sqrt{2}}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{7/5\sqrt{2}}{1/5\sqrt{2}} = 7$.
Let $\beta = \sin^{-1} \frac{4}{\sqrt{17}}$. Then $\sin \beta = \frac{4}{\sqrt{17}}$.
Since $\cos^2 \beta = 1 - \sin^2 \beta = 1 - \frac{16}{17} = \frac{1}{17}$,we have $\cos \beta = \frac{1}{\sqrt{17}}$.
Thus,$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/\sqrt{17}}{1/\sqrt{17}} = 4$.
We need to find $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Substituting the values,$\tan(\alpha - \beta) = \frac{7 - 4}{1 + (7)(4)} = \frac{3}{1 + 28} = \frac{3}{29}$.

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$. Then $\sin \alpha = \frac{3}{5}$,which implies $\tan \alpha = \frac{3}{4}$.
Let $\beta = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right)$. Then $\cos \beta = \frac{2}{\sqrt{5}}$,which implies $\tan \beta = \frac{1}{2}$.
We need to find $\tan(\alpha - 2\beta)$.
First,calculate $\tan(2\beta) = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$.
Now,use the formula $\tan(\alpha - 2\beta) = \frac{\tan \alpha - \tan 2\beta}{1 + \tan \alpha \tan 2\beta}$.
Substituting the values: $\tan(\alpha - 2\beta) = \frac{3/4 - 4/3}{1 + (3/4)(4/3)} = \frac{(9-16)/12}{1 + 1} = \frac{-7/12}{2} = -\frac{7}{24}$.

(B) Given $f(\theta) = \sin ( \tan ^{-1} ( \frac{\sin \theta}{\sqrt{\cos 2 \theta}} ) )$.
Using the identity $\tan ^{-1} x = \sin ^{-1} ( \frac{x}{\sqrt{1+x^2}} )$,we have:
$f(\theta) = \sin ( \sin ^{-1} ( \frac{\frac{\sin \theta}{\sqrt{\cos 2 \theta}}}{\sqrt{1 + \frac{\sin ^2 \theta}{\cos 2 \theta}}} ) )$
$f(\theta) = \sin ( \sin ^{-1} ( \frac{\sin \theta}{\sqrt{\cos 2 \theta + \sin ^2 \theta}} ) )$
Since $\cos 2 \theta = \cos ^2 \theta - \sin ^2 \theta$,we get:
$f(\theta) = \sin ( \sin ^{-1} ( \frac{\sin \theta}{\sqrt{\cos ^2 \theta - \sin ^2 \theta + \sin ^2 \theta}} ) )$
$f(\theta) = \sin ( \sin ^{-1} ( \frac{\sin \theta}{\sqrt{\cos ^2 \theta}} ) )$
For $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,$\cos \theta > 0$,so $\sqrt{\cos ^2 \theta} = \cos \theta$.
$f(\theta) = \sin ( \sin ^{-1} ( \frac{\sin \theta}{\cos \theta} ) ) = \sin ( \sin ^{-1} ( \tan \theta ) ) = \tan \theta$.
Now,we need to find $\frac{d}{d(\tan \theta)}(f(\theta))$.
Let $u = \tan \theta$. Then $f(\theta) = u$.
Therefore,$\frac{d}{du}(u) = 1$.

(B) We need to evaluate $\cot \left(\operatorname{cosec}^{-1} \frac{5}{3}+\tan ^{-1} \frac{2}{3}\right)$.
First,convert $\operatorname{cosec}^{-1} \frac{5}{3}$ to $\tan^{-1}$. Since $\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$,we have $\operatorname{cosec}^{-1} \frac{5}{3} = \sin^{-1} \frac{3}{5}$.
Let $\sin^{-1} \frac{3}{5} = \theta$,then $\sin \theta = \frac{3}{5}$. Thus,$\tan \theta = \frac{3}{\sqrt{5^2-3^2}} = \frac{3}{4}$. So,$\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{3}{4}$.
Now the expression becomes $\cot \left(\tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3}\right)$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$,we get:
$\tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} = \tan^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right) = \tan^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right) = \tan^{-1} \left(\frac{17/12}{6/12}\right) = \tan^{-1} \frac{17}{6}$.
Finally,$\cot \left(\tan^{-1} \frac{17}{6}\right) = \cot \left(\cot^{-1} \frac{6}{17}\right) = \frac{6}{17}$.

(C) Let $\cot ^{-1} x = \theta$,then $x = \cot \theta$.
Since $0 < x < 1$,we have $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.
The expression is $\sqrt{1+x^2} [\{x \cos \theta + \sin \theta\}^2 - 1]^{\frac{1}{2}}$.
Substituting $x = \cot \theta = \frac{\cos \theta}{\sin \theta}$,we get:
$\sqrt{1+\cot^2 \theta} [\{\frac{\cos^2 \theta}{\sin \theta} + \sin \theta\}^2 - 1]^{\frac{1}{2}}$
$= \sqrt{\operatorname{cosec}^2 \theta} [\{\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta}\}^2 - 1]^{\frac{1}{2}}$
$= \operatorname{cosec} \theta [\{\frac{1}{\sin \theta}\}^2 - 1]^{\frac{1}{2}}$
$= \operatorname{cosec} \theta \sqrt{\operatorname{cosec}^2 \theta - 1}$
$= \operatorname{cosec} \theta \sqrt{\cot^2 \theta}$
$= \operatorname{cosec} \theta \cdot \cot \theta$ (since $\cot \theta > 0$ for $0 < \theta < \frac{\pi}{2}$)
$= \frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{\cos \theta}{\sin^2 \theta} = \frac{\cos \theta}{1 - \cos^2 \theta}$.
Alternatively,using $\sqrt{1+x^2} = \sqrt{1+\cot^2 \theta} = \operatorname{cosec} \theta$:
Expression $= \operatorname{cosec} \theta \sqrt{(\cot \theta \cos \theta + \sin \theta)^2 - 1}$
$= \operatorname{cosec} \theta \sqrt{(\frac{\cos^2 \theta}{\sin \theta} + \sin \theta)^2 - 1}$
$= \operatorname{cosec} \theta \sqrt{(\frac{1}{\sin \theta})^2 - 1} = \operatorname{cosec} \theta \sqrt{\operatorname{cosec}^2 \theta - 1} = \operatorname{cosec} \theta \cot \theta$.
Since $\operatorname{cosec} \theta = \sqrt{1+x^2}$ and $\cot \theta = x$,the result is $x \sqrt{1+x^2}$.

(B) Given equation is $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$.
Since $\cos^{-1} x = \alpha$,we have $x = \cos \alpha$. Given $0 < x < 1$,we have $0 < \alpha < \frac{\pi}{2}$.
Substituting $x = \cos \alpha$ into the equation:
$\sin^{-1} (2 \cos \alpha \sqrt{1 - \cos^2 \alpha}) + \sec^{-1} (\frac{1}{2 \cos^2 \alpha - 1}) = \frac{2 \pi}{3}$
$\sin^{-1} (2 \cos \alpha \sin \alpha) + \sec^{-1} (\frac{1}{\cos 2 \alpha}) = \frac{2 \pi}{3}$
$\sin^{-1} (\sin 2 \alpha) + \cos^{-1} (\cos 2 \alpha) = \frac{2 \pi}{3}$
Since $0 < \alpha < \frac{\pi}{2}$,we have $0 < 2 \alpha < \pi$. Thus,$\sin^{-1} (\sin 2 \alpha) = 2 \alpha$ and $\cos^{-1} (\cos 2 \alpha) = 2 \alpha$.
Therefore,$2 \alpha + 2 \alpha = \frac{2 \pi}{3}$
$4 \alpha = \frac{2 \pi}{3}$
$\alpha = \frac{\pi}{6}$.