Properties of ITF Questions in English

(B) Given the equation: $\cot \frac{2x}{3} + \tan \frac{x}{3} = \csc \frac{kx}{3}$.
Using the identities $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$,we have:
$\frac{\cos(2x/3)}{\sin(2x/3)} + \frac{\sin(x/3)}{\cos(x/3)} = \frac{\cos(2x/3)\cos(x/3) + \sin(2x/3)\sin(x/3)}{\sin(2x/3)\cos(x/3)}$.
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$,the numerator becomes $\cos(2x/3 - x/3) = \cos(x/3)$.
So,the expression is $\frac{\cos(x/3)}{\sin(2x/3)\cos(x/3)} = \frac{1}{\sin(2x/3)} = \csc \frac{2x}{3}$.
Comparing this with $\csc \frac{kx}{3}$,we get $k = 2$.
We need to find $\tan^{-1}(\tan 2)$.
Since $2$ radians is in the second quadrant $(\pi/2 < 2 < \pi)$,the principal value of $\tan^{-1}(\tan 2)$ is $2 - \pi$.

(D) Let $\sqrt{x} = \sin \theta$,which implies $x = \sin^2 \theta$.
Since $0 < x < \frac{1}{2}$,we have $0 < \sin^2 \theta < \frac{1}{2}$,so $0 < \sin \theta < \frac{1}{\sqrt{2}}$,which means $0 < \theta < \frac{\pi}{4}$.
Then,$\sqrt{1-x} = \sqrt{1-\sin^2 \theta} = \cos \theta$.
Also,$2\sqrt{x(1-x)} = 2\sin \theta \cos \theta = \sin 2\theta$.
Substituting these into the expression for $y$:
$y = 2\sin^{-1}(\cos \theta) + \sin^{-1}(\sin 2\theta)$.
Since $0 < \theta < \frac{\pi}{4}$,we have $\frac{\pi}{4} < 2\theta < \frac{\pi}{2}$,so $\sin^{-1}(\sin 2\theta) = 2\theta$.
Also,$\sin^{-1}(\cos \theta) = \sin^{-1}(\sin(\frac{\pi}{2} - \theta)) = \frac{\pi}{2} - \theta$.
Thus,$y = 2(\frac{\pi}{2} - \theta) + 2\theta = \pi - 2\theta + 2\theta = \pi$.
Since $y = \pi$ is a constant,$\frac{dy}{dx} = 0$.

(B) Let $S_n = \sum_{r=1}^n \tan^{-1} \left( \frac{4}{4r^2 + 3} \right)$.
We can rewrite the argument of the $\tan^{-1}$ function as:
$\frac{4}{4r^2 + 3} = \frac{(r + 1/2) - (r - 1/2)}{1 + (r + 1/2)(r - 1/2)} = \frac{4}{4r^2 + 1 - 1 + 3} = \frac{4}{4r^2 + 3}$.
Thus,$\tan^{-1} \left( \frac{4}{4r^2 + 3} \right) = \tan^{-1} \left( r + \frac{1}{2} \right) - \tan^{-1} \left( r - \frac{1}{2} \right)$.
This is a telescoping sum:
$S_n = \sum_{r=1}^n \left[ \tan^{-1} \left( r + \frac{1}{2} \right) - \tan^{-1} \left( r - \frac{1}{2} \right) \right] = \tan^{-1} \left( n + \frac{1}{2} \right) - \tan^{-1} \left( \frac{1}{2} \right)$.
As $n \to \infty$,$\tan^{-1} \left( n + \frac{1}{2} \right) \to \frac{\pi}{2}$.
So,the sum is $\frac{\pi}{2} - \tan^{-1} \left( \frac{1}{2} \right) = \cot^{-1} \left( \frac{1}{2} \right) = \tan^{-1} (2)$.
Finally,$\cot \left( \tan^{-1} (2) \right) = \frac{1}{2}$.

(D) We know that $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ and $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.
Given $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Using the substitution $x = \tan \theta$,we have $\frac{2x}{1+x^2} = \sin(2\theta)$ and $\frac{1-x^2}{1+x^2} = \cos(2\theta)$.
For $x \ge 1$,$f(x) = \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{2x}{1+x^2}\right)\right) + \left(\frac{\pi}{2} - \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$.
For $x \ge 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) - \pi$ is not correct; rather,for $x > 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}(x)$ and $\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x)$.
Thus,$f(x) = (\pi - 2\tan^{-1}(x)) + (2\tan^{-1}(x)) = \pi$ for $x \ge 1$.
Therefore,$f(1) = \pi$ and $f(2) = \pi$.
Hence,$f(1) + f(2) = \pi + \pi = 2\pi$.

(C) Given that $\log_{\pi}x > 0$. Since the base $\pi > 1$,this implies $x > 1$.
We know the identity for $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$:
For $x > 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}x$.
Substituting this into the expression:
$\log_{\pi}\left( \sin^{-1}\frac{2x}{1+x^2} + 2\tan^{-1}x \right) = \log_{\pi}\left( (\pi - 2\tan^{-1}x) + 2\tan^{-1}x \right)$.
Simplifying the expression inside the logarithm:
$\log_{\pi}(\pi) = 1$.

(B) Let $u = \tan^{-1} \left( \frac{x}{\sqrt{1 - x^2}} \right)$ and $v = \sin^{-1}(x)$.
Substitute $x = \sin(\theta)$,where $\theta = \sin^{-1}(x)$.
Then,$u = \tan^{-1} \left( \frac{\sin(\theta)}{\sqrt{1 - \sin^2(\theta)}} \right) = \tan^{-1} \left( \frac{\sin(\theta)}{\cos(\theta)} \right) = \tan^{-1}(\tan(\theta)) = \theta$.
Since $v = \sin^{-1}(x) = \theta$,we have $u = v$.
Therefore,the derivative of $u$ with respect to $v$ is $\frac{du}{dv} = \frac{d}{dv}(v) = 1$.

(A) The general term of the series is $T_r = \tan^{-1}\left(\frac{1}{x^2 + (2r-1)x + (r^2-r+1)}\right)$ for $r = 1, 2, \dots, n$.
This can be written as $T_r = \tan^{-1}\left(\frac{(x+r) - (x+r-1)}{1 + (x+r)(x+r-1)}\right)$.
Using the formula $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we get $T_r = \tan^{-1}(x+r) - \tan^{-1}(x+r-1)$.
Summing up to $n$ terms:
$y = [\tan^{-1}(x+1) - \tan^{-1} x] + [\tan^{-1}(x+2) - \tan^{-1}(x+1)] + \dots + [\tan^{-1}(x+n) - \tan^{-1}(x+n-1)]$.
This is a telescoping series,so $y = \tan^{-1}(x+n) - \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{1}{1 + (x+n)^2} - \frac{1}{1 + x^2}$.

(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:
$\tan^{-1}\left(\frac{2 \cos x}{\sin^2 x}\right) = \tan^{-1}\left(\frac{1}{\sin^2 x}\right)$.
Equating the arguments:
$\frac{2 \cos x}{\sin^2 x} = \frac{1}{\sin^2 x}$.
Since $\sin^2 x \neq 0$ (as $\csc^2 x$ is defined),we can cancel $\sin^2 x$ from both sides:
$2 \cos x = 1$.
$\cos x = \frac{1}{2}$.
Thus,$x = \frac{\pi}{3}$.

(D) We are given the expression $f(x) = 2 \tan^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$.
We know the identity for $2 \tan^{-1} x$ in terms of $\sin^{-1}$ is:
$2 \tan^{-1} x = \begin{cases} \sin^{-1} (\frac{2x}{1+x^2}) & \text{if } |x| \leq 1 \\ \pi - \sin^{-1} (\frac{2x}{1+x^2}) & \text{if } x > 1 \\ -\pi - \sin^{-1} (\frac{2x}{1+x^2}) & \text{if } x < -1 \end{cases}$
For $x = 1$,$2 \tan^{-1} (1) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$ and $\sin^{-1} (\frac{2(1)}{1+1^2}) = \sin^{-1}(1) = \frac{\pi}{2}$. Thus,$f(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
For $x > 1$,we have $\sin^{-1} (\frac{2x}{1+x^2}) = \pi - 2 \tan^{-1} x$.
Substituting this into the expression:
$f(x) = 2 \tan^{-1} x + (\pi - 2 \tan^{-1} x) = \pi$.
Therefore,for all $x \geq 1$,the value of the expression is $\pi$.

(B) Given,$\alpha = \tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)$.
Since $\frac{5 \pi}{4} = \pi + \frac{\pi}{4}$,we have $\tan \frac{5 \pi}{4} = \tan \frac{\pi}{4}$.
Thus,$\alpha = \tan ^{-1}\left(\tan \frac{\pi}{4}\right) = \frac{\pi}{4}$.
This implies $4 \alpha = \pi$ $(1)$.
Given,$\beta = \tan ^{-1}\left(-\tan \frac{2 \pi}{3}\right)$.
Since $\tan \frac{2 \pi}{3} = \tan \left(\pi - \frac{\pi}{3}\right) = -\tan \frac{\pi}{3}$,we have $-\tan \frac{2 \pi}{3} = \tan \frac{\pi}{3}$.
Thus,$\beta = \tan ^{-1}\left(\tan \frac{\pi}{3}\right) = \frac{\pi}{3}$.
This implies $3 \beta = \pi$ $(2)$.
From equations $(1)$ and $(2)$,we have $4 \alpha = \pi$ and $3 \beta = \pi$.
Therefore,$4 \alpha = 3 \beta$,which means $4 \alpha - 3 \beta = 0$.

(C) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$. We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.
Using the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$,we get:
$f(x) = (\sin^{-1} x + \cos^{-1} x)^3 - 3(\sin^{-1} x)(\cos^{-1} x)(\sin^{-1} x + \cos^{-1} x)$
$f(x) = (\frac{\pi}{2})^3 - 3(\sin^{-1} x)(\frac{\pi}{2} - \sin^{-1} x)(\frac{\pi}{2})$
$f(x) = \frac{\pi^3}{8} - \frac{3\pi}{2}(\sin^{-1} x)(\frac{\pi}{2} - \sin^{-1} x)$
$f(x) = \frac{\pi^3}{8} - \frac{3\pi^2}{4}(\sin^{-1} x) + \frac{3\pi}{2}(\sin^{-1} x)^2$
Completing the square for the quadratic in $\sin^{-1} x$:
$f(x) = \frac{3\pi}{2} [(\sin^{-1} x)^2 - \frac{\pi}{2}(\sin^{-1} x) + \frac{\pi^2}{16} - \frac{\pi^2}{16}] + \frac{\pi^3}{8}$
$f(x) = \frac{3\pi}{2}(\sin^{-1} x - \frac{\pi}{4})^2 - \frac{3\pi^3}{32} + \frac{4\pi^3}{32} = \frac{3\pi}{2}(\sin^{-1} x - \frac{\pi}{4})^2 + \frac{\pi^3}{32}$
Since $\sin^{-1} x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $\sin^{-1} x = \frac{\pi}{4}$,which is $\frac{\pi^3}{32}$.
The maximum value occurs at the boundary $\sin^{-1} x = -\frac{\pi}{2}$:
$f(-\frac{\pi}{2}) = \frac{3\pi}{2}(-\frac{\pi}{2} - \frac{\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(-\frac{3\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(\frac{9\pi^2}{16}) + \frac{\pi^3}{32} = \frac{27\pi^3}{32} + \frac{\pi^3}{32} = \frac{28\pi^3}{32} = \frac{7\pi^3}{8}$.

(D) We know that the range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.
Given the interval $\pi \leq x \leq 2 \pi$.
We use the property $\cos(2 \pi - x) = \cos x$.
Since $\pi \leq x \leq 2 \pi$,we have $-\pi \geq -x \geq -2 \pi$.
Adding $2 \pi$ to all sides,we get $2 \pi - \pi \geq 2 \pi - x \geq 2 \pi - 2 \pi$,which simplifies to $\pi \geq 2 \pi - x \geq 0$.
Thus,$0 \leq 2 \pi - x \leq \pi$,which lies within the principal range $[0, \pi]$.
Therefore,$\cos^{-1}(\cos x) = \cos^{-1}(\cos(2 \pi - x)) = 2 \pi - x$.

(D) We are given the equation: $\sin(\cot^{-1}(1 + x)) = \cos(\tan^{-1} x)$.
First,simplify the right-hand side: Let $\tan^{-1} x = \theta$,then $x = \tan \theta$.
Since $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1 + \tan^2 \theta}}$,we have $\cos(\tan^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$.
Next,simplify the left-hand side: Let $\cot^{-1}(1 + x) = \phi$,then $1 + x = \cot \phi$.
Since $\sin \phi = \frac{1}{\csc \phi} = \frac{1}{\sqrt{1 + \cot^2 \phi}}$,we have $\sin(\cot^{-1}(1 + x)) = \frac{1}{\sqrt{1 + (1 + x)^2}}$.
Equating both sides: $\frac{1}{\sqrt{1 + (1 + x)^2}} = \frac{1}{\sqrt{1 + x^2}}$.
Squaring both sides: $1 + (1 + x)^2 = 1 + x^2$.
$(1 + x)^2 = x^2$.
$1 + 2x + x^2 = x^2$.
$1 + 2x = 0$.
$x = -\frac{1}{2}$.

(C) Given the inequality: ${\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6}$.
Since the function $f(x) = {\cot ^{ - 1}}x$ is a strictly decreasing function,the inequality sign reverses when we take the cotangent of both sides:
$\frac{n}{\pi } < \cot \left( \frac{\pi }{6} \right)$
We know that $\cot \left( \frac{\pi }{6} \right) = \sqrt{3}$.
So,$\frac{n}{\pi } < \sqrt{3}$.
Multiplying both sides by $\pi$:
$n < \pi \sqrt{3}$.
Using the approximation $\pi \approx 3.14159$ and $\sqrt{3} \approx 1.732$:
$n < 3.14159 \times 1.732 \approx 5.44$.
Since $n \in N$ (natural numbers),the maximum integer value of $n$ satisfying $n < 5.44$ is $5$.

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

(A) Let $\lambda = \cot^{-1}(1+x)$,then $\cot \lambda = 1+x$. From the right-angled triangle with base $(1+x)$ and perpendicular $1$,the hypotenuse is $\sqrt{(1+x)^2 + 1^2} = \sqrt{x^2 + 2x + 2}$. Thus,$\sin \lambda = \frac{1}{\sqrt{x^2 + 2x + 2}}$.
Let $\beta = \tan^{-1}x$,then $\tan \beta = x$. From the right-angled triangle with perpendicular $x$ and base $1$,the hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$. Thus,$\cos \beta = \frac{1}{\sqrt{x^2 + 1}}$.
Given the equation $\sin \lambda = \cos \beta$,we have:
$\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}}$
Squaring both sides:
$x^2 + 2x + 2 = x^2 + 1$
Subtracting $x^2$ from both sides:
$2x + 2 = 1$
$2x = -1$
$x = -\frac{1}{2}$

(C) Given the function $f(x) = 2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$ for $x > 1$.
We know that for $x > 1$,the formula for $\sin^{-1} \left( \frac{2x}{1 + x^2} \right)$ is $\pi - 2 \tan^{-1} x$.
Substituting this into the expression for $f(x)$:
$f(x) = 2 \tan^{-1} x + (\pi - 2 \tan^{-1} x)$
$f(x) = \pi$.
Since $f(x)$ is a constant function for $x > 1$,$f(5) = \pi$.

(D) For Statement $II$: We know that $\sin^{-1} x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Subtracting $\frac{\pi}{4}$ from all sides,we get $-\frac{3\pi}{4} \le \sin^{-1} x - \frac{\pi}{4} \le \frac{\pi}{4}$.
Squaring the inequality,we get $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}$. Thus,Statement $II$ is true.
For Statement $I$: Let $u = \sin^{-1} x$. Then $\cos^{-1} x = \frac{\pi}{2} - u$. The equation becomes $u^3 + (\frac{\pi}{2} - u)^3 = a\pi^3$.
Expanding this,we get $u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = a\pi^3$.
$\frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8} - a\pi^3 = 0$.
Dividing by $\frac{3\pi}{2}$,we get $u^2 - \frac{\pi}{2}u + \frac{\pi^2}{12} - \frac{2a\pi^2}{3} = 0$.
Completing the square: $(u - \frac{\pi}{4})^2 - \frac{\pi^2}{16} + \frac{\pi^2}{12} - \frac{2a\pi^2}{3} = 0$.
$(u - \frac{\pi}{4})^2 = \frac{2a\pi^2}{3} - \frac{\pi^2}{48} = \frac{\pi^2}{48}(32a - 1)$.
Since $0 \le (u - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}$,we have $0 \le \frac{\pi^2}{48}(32a - 1) \le \frac{9\pi^2}{16}$.
$0 \le 32a - 1 \le 27$,which implies $\frac{1}{32} \le a \le \frac{7}{8}$.
Since the equation has a solution only for $a \in [\frac{1}{32}, \frac{7}{8}]$,the statement that it has a solution for all $a \ge \frac{1}{32}$ is false.

(D) Given equation is $\sin^{-1} x = 2 \tan^{-1} x$.
Let $\tan^{-1} x = \theta$,then $x = \tan \theta$.
Substituting this into the equation,we get $\sin^{-1}(\tan \theta) = 2\theta$.
Taking $\sin$ on both sides,$\tan \theta = \sin(2\theta)$.
Using the identity $\sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we have $\tan \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$.
This implies $\tan \theta (1 - \frac{2}{1 + \tan^2 \theta}) = 0$.
So,$\tan \theta = 0$ or $1 + \tan^2 \theta = 2$,which means $\tan^2 \theta = 1$.
If $\tan \theta = 0$,then $x = 0$.
If $\tan^2 \theta = 1$,then $\tan \theta = 1$ or $\tan \theta = -1$,so $x = 1$ or $x = -1$.
Checking these values in the original equation:
For $x = 0$: $\sin^{-1}(0) = 0$ and $2 \tan^{-1}(0) = 0$. (Valid)
For $x = 1$: $\sin^{-1}(1) = \frac{\pi}{2}$ and $2 \tan^{-1}(1) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$. (Valid)
For $x = -1$: $\sin^{-1}(-1) = -\frac{\pi}{2}$ and $2 \tan^{-1}(-1) = 2(-\frac{\pi}{4}) = -\frac{\pi}{2}$. (Valid)
Thus,there are $3$ solutions: $x \in \{-1, 0, 1\}$.

(A) We use the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$.
Each term in the series is of the form $\tan^{-1} \left( \frac{1}{1 + k(k+1)} \right) = \tan^{-1}(k+1) - \tan^{-1}(k)$.
Let $f(k) = \tan^{-1}(k+1) - \tan^{-1}(k)$.
The given sum is $S = \sum_{k=n}^{n+19} \tan^{-1} \left( \frac{1}{1 + k(k+1)} \right)$.
$S = \sum_{k=n}^{n+19} (\tan^{-1}(k+1) - \tan^{-1}(k))$.
This is a telescoping sum:
$S = (\tan^{-1}(n+1) - \tan^{-1}(n)) + (\tan^{-1}(n+2) - \tan^{-1}(n+1)) + \dots + (\tan^{-1}(n+20) - \tan^{-1}(n+19))$.
$S = \tan^{-1}(n+20) - \tan^{-1}(n)$.
Using the formula $\tan S = \tan(\tan^{-1}(n+20) - \tan^{-1}(n)) = \frac{(n+20) - n}{1 + (n+20)n}$.
$\tan S = \frac{20}{1 + n^2 + 20n} = \frac{20}{n^2 + 20n + 1}$.

(B) Given the inequality $\sin^{-1} x > \cos^{-1} x$ for $x \in (0, 1)$.
We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.
Substituting this into the inequality,we get:
$\sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x$
$2 \sin^{-1} x > \frac{\pi}{2}$
$\sin^{-1} x > \frac{\pi}{4}$
Since the sine function is strictly increasing on the interval $[0, 1]$,we apply the sine function to both sides:
$x > \sin\left( \frac{\pi}{4} \right)$
$x > \frac{1}{\sqrt{2}}$
Given the domain $x \in (0, 1)$,the solution set is $x \in \left( \frac{1}{\sqrt{2}}, 1 \right)$.

(A) Given $\cos^{-1}\left(\frac{2}{3x}\right) + \cos^{-1}\left(\frac{3}{4x}\right) = \frac{\pi}{2}$.
We know that $\cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}\left(AB - \sqrt{1-A^2}\sqrt{1-B^2}\right)$.
So,$\cos^{-1}\left(\frac{2}{3x} \cdot \frac{3}{4x} - \sqrt{1-\frac{4}{9x^2}}\sqrt{1-\frac{9}{16x^2}}\right) = \frac{\pi}{2}$.
Taking $\cos$ on both sides,$\frac{6}{12x^2} - \sqrt{\frac{9x^2-4}{9x^2}}\sqrt{\frac{16x^2-9}{16x^2}} = \cos\left(\frac{\pi}{2}\right) = 0$.
$\frac{1}{2x^2} = \frac{\sqrt{9x^2-4}\sqrt{16x^2-9}}{12x^2}$.
Multiplying by $12x^2$,we get $6 = \sqrt{9x^2-4}\sqrt{16x^2-9}$.
Squaring both sides,$36 = (9x^2-4)(16x^2-9) = 144x^4 - 81x^2 - 64x^2 + 36$.
$144x^4 - 145x^2 = 0$.
$x^2(144x^2 - 145) = 0$.
Since $x > \frac{3}{4}$,$x^2 \neq 0$,so $144x^2 = 145$,which gives $x = \frac{\sqrt{145}}{12}$.

(A) We know that the principal value range for $\sin^{-1}(\sin \theta)$ is $[-\pi/2, \pi/2]$.
Since $3\pi \approx 9.42$ and $10$ lies in the interval $(2\pi, 3\pi)$,we use the property $\sin^{-1}(\sin \theta) = \theta - 3\pi$ for $\theta \in [2\pi, 3\pi]$.
Thus,$x = \sin^{-1}(\sin 10) = 10 - 3\pi$.
However,looking at the graph provided,the value is $3\pi - 10$.
Let's re-evaluate: $10$ radians is in the third quadrant ($3\pi < 10 < 3.5\pi$ is false,$3\pi \approx 9.42$ and $3.5\pi \approx 10.99$).
Actually,$3\pi < 10 < 3.5\pi$,so $\sin(10) = \sin(10 - 3\pi) = -sin(3\pi - 10) = sin(3\pi - 10)$.
Since $3\pi - 10 \in [-\pi/2, \pi/2]$,$x = 3\pi - 10$.
For $y = \cos^{-1}(\cos 10)$,the range is $[0, \pi]$.
Since $3\pi < 10 < 4\pi$,we use $\cos^{-1}(\cos \theta) = 4\pi - \theta$.
Thus,$y = 4\pi - 10$.
Therefore,$y - x = (4\pi - 10) - (3\pi - 10) = \pi$.

(A) We know that $\sum_{p=1}^n 2p = 2 \times \frac{n(n+1)}{2} = n^2 + n$.
So,the expression inside the summation is $\cot^{-1}(1 + n^2 + n) = \tan^{-1}\left(\frac{1}{1 + n(n+1)}\right)$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$,we can write:
$\tan^{-1}\left(\frac{(n+1) - n}{1 + n(n+1)}\right) = \tan^{-1}(n+1) - \tan^{-1}(n)$.
Now,the sum becomes:
$\sum_{n=1}^{19} (\tan^{-1}(n+1) - \tan^{-1}(n)) = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1} 20 - \tan^{-1} 19)$.
This is a telescoping sum,which simplifies to $\tan^{-1} 20 - \tan^{-1} 1$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$:
$\tan^{-1} 20 - \tan^{-1} 1 = \tan^{-1}\left(\frac{20-1}{1+20 \times 1}\right) = \tan^{-1}\left(\frac{19}{21}\right)$.
Finally,we need to find $\cot(\tan^{-1}(19/21)) = \cot(\cot^{-1}(21/19)) = \frac{21}{19}$.

(A) Given $\cos \alpha = \frac{3}{5}$. Since $\cos^2 \alpha + \sin^2 \alpha = 1$,we have $\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{3/5} = \frac{4}{3}$.
We are given $\tan \beta = \frac{1}{3}$.
Using the formula $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$,we get:
$\tan(\alpha - \beta) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + (\frac{4}{3})(\frac{1}{3})} = \frac{1}{1 + \frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13}$.
Now,if $\tan(\alpha - \beta) = \frac{9}{13}$,then $\sin(\alpha - \beta) = \frac{9}{\sqrt{9^2 + 13^2}} = \frac{9}{\sqrt{81 + 169}} = \frac{9}{\sqrt{250}} = \frac{9}{5\sqrt{10}}$.
Therefore,$\alpha - \beta = \sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)$.

(D) Let $\alpha = \sin^{-1} \left( \frac{12}{13} \right)$ and $\beta = \sin^{-1} \left( \frac{3}{5} \right)$.
Then $\sin \alpha = \frac{12}{13} \implies \cos \alpha = \sqrt{1 - \left( \frac{12}{13} \right)^2} = \frac{5}{13}$.
And $\sin \beta = \frac{3}{5} \implies \cos \beta = \sqrt{1 - \left( \frac{3}{5} \right)^2} = \frac{4}{5}$.
Using the formula $\sin^{-1} x - \sin^{-1} y = \sin^{-1} (x \sqrt{1 - y^2} - y \sqrt{1 - x^2})$:
$\sin^{-1} \left( \frac{12}{13} \right) - \sin^{-1} \left( \frac{3}{5} \right) = \sin^{-1} \left( \frac{12}{13} \times \frac{4}{5} - \frac{3}{5} \times \frac{5}{13} \right)$
$= \sin^{-1} \left( \frac{48}{65} - \frac{15}{65} \right) = \sin^{-1} \left( \frac{33}{65} \right)$.
Since $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we have $\sin^{-1} \left( \frac{33}{65} \right) = \frac{\pi}{2} - \cos^{-1} \left( \frac{33}{65} \right)$.
Alternatively,using $\cos^{-1} x = \sin^{-1} \sqrt{1 - x^2}$,we have $\sin^{-1} \left( \frac{33}{65} \right) = \cos^{-1} \sqrt{1 - \left( \frac{33}{65} \right)^2} = \cos^{-1} \left( \frac{56}{65} \right) = \frac{\pi}{2} - \sin^{-1} \left( \frac{56}{65} \right)$.
Thus,the correct option is $D$.

(N/A) Let $x = \sin \theta$. Then $\theta = \sin^{-1} x$.
Since $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$,we have $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$,which implies $-\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2}$.
Now,consider the expression $\sin^{-1}(2x\sqrt{1-x^2})$.
Substituting $x = \sin \theta$,we get $\sin^{-1}(2\sin \theta \sqrt{1-\sin^2 \theta})$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we have $\sin^{-1}(2\sin \theta \sqrt{\cos^2 \theta}) = \sin^{-1}(2\sin \theta |\cos \theta|)$.
Since $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$,$\cos \theta \geq 0$,so $|\cos \theta| = \cos \theta$.
Thus,the expression becomes $\sin^{-1}(2\sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta)$.
Since $2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$\sin^{-1}(\sin 2\theta) = 2\theta$.
Substituting back $\theta = \sin^{-1} x$,we get $2\sin^{-1} x$.

Let $x = \cos \theta$. Then $\theta = \cos ^{-1} x$.
Given the range $\frac{1}{\sqrt{2}} \leq x \leq 1$,we have $\cos \frac{\pi}{4} \leq \cos \theta \leq \cos 0$,which implies $0 \leq \theta \leq \frac{\pi}{4}$.
Thus,$0 \leq 2\theta \leq \frac{\pi}{2}$.
Now,substitute $x = \cos \theta$ into the expression:
$\sin ^{-1}(2 x \sqrt{1-x^{2}}) = \sin ^{-1}(2 \cos \theta \sqrt{1-\cos ^{2} \theta})$
$= \sin ^{-1}(2 \cos \theta \sin \theta)$
$= \sin ^{-1}(\sin 2 \theta)$
Since $0 \leq 2\theta \leq \frac{\pi}{2}$,$\sin ^{-1}(\sin 2 \theta) = 2 \theta$.
Substituting back $\theta = \cos ^{-1} x$,we get $2 \cos ^{-1} x$.
Hence,$\sin ^{-1}(2 x \sqrt{1-x^{2}}) = 2 \cos ^{-1} x$.

(A) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
Given $L.H.S. = \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{2}{11}$.
Applying the formula with $x = \frac{1}{2}$ and $y = \frac{2}{11}$:
$L.H.S. = \tan ^{-1} \left( \frac{\frac{1}{2} + \frac{2}{11}}{1 - (\frac{1}{2} \times \frac{2}{11})} \right)$.
Simplifying the numerator: $\frac{1}{2} + \frac{2}{11} = \frac{11 + 4}{22} = \frac{15}{22}$.
Simplifying the denominator: $1 - \frac{2}{22} = 1 - \frac{1}{11} = \frac{10}{11}$.
Thus,$L.H.S. = \tan ^{-1} \left( \frac{15/22}{10/11} \right) = \tan ^{-1} \left( \frac{15}{22} \times \frac{11}{10} \right)$.
$L.H.S. = \tan ^{-1} \left( \frac{15}{2 \times 10} \right) = \tan ^{-1} \left( \frac{15}{20} \right) = \tan ^{-1} \frac{3}{4}$.
Since $L.H.S. = R.H.S.$,the identity is proved.

(D) Given expression: $\tan ^{-1} \left( \frac{\cos x}{1-\sin x} \right)$
Using trigonometric identities $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$,$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,and $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$:
$= \tan ^{-1} \left[ \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}} \right]$
$= \tan ^{-1} \left[ \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} \right]$
$= \tan ^{-1} \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right)$
Dividing numerator and denominator by $\cos \frac{x}{2}$:
$= \tan ^{-1} \left( \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right) = \tan ^{-1} \left[ \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right]$
$= \frac{\pi}{4} + \frac{x}{2}$

(A) Let $x = \tan \theta$. Then $\theta = \tan ^{-1} x$.
Consider the $R.H.S.$:
$\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) = \tan ^{-1}\left(\frac{3 \tan \theta - \tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$
Using the trigonometric identity $\tan 3\theta = \frac{3 \tan \theta - \tan ^{3} \theta}{1-3 \tan ^{2} \theta}$,we get:
$= \tan ^{-1}(\tan 3 \theta) = 3 \theta$
Since $3\theta = \theta + 2\theta$,we can write:
$= \tan ^{-1} x + 2 \tan ^{-1} x$
Using the formula $2 \tan ^{-1} x = \tan ^{-1} \left(\frac{2x}{1-x^2}\right)$ for $|x|<1$:
$= \tan ^{-1} x + \tan ^{-1} \left(\frac{2x}{1-x^2}\right) = L.H.S.$
Hence,the identity is proved.

(C) We know the identity for inverse trigonometric functions: $\sec ^{-1} x + \csc ^{-1} x = \frac{\pi}{2}$ for $|x| \geq 1$.
Substituting this into the given expression:
$\cos \left(\sec ^{-1} x + \csc ^{-1} x\right) = \cos \left(\frac{\pi}{2}\right)$.
Since $\cos \left(\frac{\pi}{2}\right) = 0$,the value of the expression is $0$.

(N/A) To prove $3 \sin ^{-1} x = \sin ^{-1}(3 x - 4 x^{3})$,where $x \in [-\frac{1}{2}, \frac{1}{2}]$.
Let $x = \sin \theta$. Then,$\theta = \sin ^{-1} x$.
Since $x \in [-\frac{1}{2}, \frac{1}{2}]$,we have $\sin \theta \in [-\frac{1}{2}, \frac{1}{2}]$,which implies $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.
Multiplying by $3$,we get $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Now,consider the $R$.$H$.$S$.:
$\sin ^{-1}(3 x - 4 x^{3}) = \sin ^{-1}(3 \sin \theta - 4 \sin ^{3} \theta)$
Using the trigonometric identity $\sin 3\theta = 3 \sin \theta - 4 \sin ^{3} \theta$,we get:
$= \sin ^{-1}(\sin 3 \theta)$
Since $3\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we can use the property $\sin ^{-1}(\sin \alpha) = \alpha$ for $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
$= 3 \theta$
$= 3 \sin ^{-1} x = \text{L.H.S.}$
Hence,the identity is proved.

To prove $3 \cos ^{-1} x = \cos ^{-1} (4 x^{3} - 3 x)$ for $x \in [\frac{1}{2}, 1]$.
Let $x = \cos \theta$. Then $\theta = \cos ^{-1} x$.
Since $x \in [\frac{1}{2}, 1]$,we have $\cos \theta \in [\frac{1}{2}, 1]$,which implies $\theta \in [0, \frac{\pi}{3}]$.
Multiplying by $3$,we get $3\theta \in [0, \pi]$.
Now,consider the $R.H.S$:
$\cos ^{-1} (4 x^{3} - 3 x)$
$= \cos ^{-1} (4 \cos ^{3} \theta - 3 \cos \theta)$
Using the trigonometric identity $\cos 3\theta = 4 \cos ^{3} \theta - 3 \cos \theta$,we get:
$= \cos ^{-1} (\cos 3\theta)$
Since $3\theta \in [0, \pi]$,we have $\cos ^{-1} (\cos 3\theta) = 3\theta$.
$= 3 \cos ^{-1} x = L.H.S$.
Hence,the identity is proved.

We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
$L.H.S. = \tan ^{-1} \left( \frac{2}{11} \right) + \tan ^{-1} \left( \frac{7}{24} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{11} + \frac{7}{24}}{1 - (\frac{2}{11} \times \frac{7}{24})} \right)$
$= \tan ^{-1} \left( \frac{\frac{48 + 77}{264}}{\frac{264 - 14}{264}} \right)$
$= \tan ^{-1} \left( \frac{125}{250} \right)$
$= \tan ^{-1} \left( \frac{1}{2} \right) = R.H.S.$

$L$.$H$.$S$ $= 2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7}$
Using the formula $2 \tan ^{-1} x = \tan ^{-1} \frac{2x}{1-x^2}$,we get:
$= \tan ^{-1} \left( \frac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2} \right) + \tan ^{-1} \frac{1}{7}$
$= \tan ^{-1} \left( \frac{1}{1 - \frac{1}{4}} \right) + \tan ^{-1} \frac{1}{7}$
$= \tan ^{-1} \left( \frac{1}{\frac{3}{4}} \right) + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{1}{7}$
Now,using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$:
$= \tan ^{-1} \left( \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \cdot \frac{1}{7}} \right)$
$= \tan ^{-1} \left( \frac{\frac{28+3}{21}}{\frac{21-4}{21}} \right) = \tan ^{-1} \left( \frac{31}{17} \right) = R.H.S$.

(D) Let the given expression be $y = \tan ^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right)$.
Substitute $x = \tan \theta$,which implies $\theta = \tan ^{-1} x$.
Then,the expression becomes:
$y = \tan ^{-1} \left( \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta} \right)$
Using the identity $1 + \tan^2 \theta = \sec^2 \theta$,we get:
$y = \tan ^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right)$
Converting to $\sin \theta$ and $\cos \theta$:
$y = \tan ^{-1} \left( \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} \right) = \tan ^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right)$
Using half-angle formulas $1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right)$ and $\sin \theta = 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$:
$y = \tan ^{-1} \left( \frac{2 \sin^2 \left( \frac{\theta}{2} \right)}{2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)} \right)$
$y = \tan ^{-1} \left( \tan \left( \frac{\theta}{2} \right) \right) = \frac{\theta}{2}$
Substituting back $\theta = \tan ^{-1} x$,we get:
$y = \frac{1}{2} \tan ^{-1} x$.

(A) Given expression: $\tan ^{-1} \left(\frac{1}{\sqrt{x^{2}-1}}\right), |x|>1$
Let $x = \sec \theta$,then $\theta = \sec^{-1} x$.
Since $|x| > 1$,$\theta \in (0, \pi/2) \cup (\pi/2, \pi)$.
Substituting $x = \sec \theta$ into the expression:
$= \tan ^{-1} \left(\frac{1}{\sqrt{\sec^{2} \theta - 1}}\right)$
$= \tan ^{-1} \left(\frac{1}{\sqrt{\tan^{2} \theta}}\right)$
$= \tan ^{-1} \left(\frac{1}{|\tan \theta|}\right)$
For $x > 1$,$\theta \in (0, \pi/2)$,so $\tan \theta > 0$,thus $|\tan \theta| = \tan \theta$.
$= \tan ^{-1} \left(\frac{1}{\tan \theta}\right) = \tan ^{-1} (\cot \theta)$
$= \tan ^{-1} \left(\tan \left(\frac{\pi}{2} - \theta\right)\right)$
$= \frac{\pi}{2} - \theta$
$= \frac{\pi}{2} - \sec^{-1} x$.

(A) Consider the expression $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)$.
Let $x = a \tan \theta$,which implies $\frac{x}{a} = \tan \theta$ or $\theta = \tan ^{-1}\left(\frac{x}{a}\right)$.
Substituting $x = a \tan \theta$ into the expression:
$\tan ^{-1}\left(\frac{3 a^{2} (a \tan \theta) - (a \tan \theta)^{3}}{a^{3} - 3 a (a \tan \theta)^{2}}\right)$
$= \tan ^{-1}\left(\frac{3 a^{3} \tan \theta - a^{3} \tan ^{3} \theta}{a^{3} - 3 a^{3} \tan ^{2} \theta}\right)$
$= \tan ^{-1}\left(\frac{a^{3}(3 \tan \theta - \tan ^{3} \theta)}{a^{3}(1 - 3 \tan ^{2} \theta)}\right)$
$= \tan ^{-1}\left(\frac{3 \tan \theta - \tan ^{3} \theta}{1 - 3 \tan ^{2} \theta}\right)$
Using the trigonometric identity $\tan 3\theta = \frac{3 \tan \theta - \tan ^{3} \theta}{1 - 3 \tan ^{2} \theta}$,we get:
$= \tan ^{-1}(\tan 3 \theta)$
Since $\frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}$,we have $\frac{-1}{\sqrt{3}} \leq \tan \theta \leq \frac{1}{\sqrt{3}}$,which means $-\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}$,so $-\frac{\pi}{2} \leq 3\theta \leq \frac{\pi}{2}$.
Thus,$\tan ^{-1}(\tan 3 \theta) = 3 \theta = 3 \tan ^{-1} \frac{x}{a}$.

(B) We know that for any real number $a$,the sum of the inverse trigonometric functions is given by the identity: $\tan ^{-1} a + \cot ^{-1} a = \frac{\pi}{2}$.
Substituting this identity into the given expression:
$\cot \left(\tan ^{-1} a + \cot ^{-1} a\right) = \cot \left(\frac{\pi}{2}\right)$.
Since $\cot \left(\frac{\pi}{2}\right) = 0$,the value of the expression is $0$.

(D) Let $x = \tan \alpha$. Then $\alpha = \tan^{-1} x$.
$\sin^{-1} \left( \frac{2x}{1+x^2} \right) = \sin^{-1} \left( \frac{2 \tan \alpha}{1+\tan^2 \alpha} \right) = \sin^{-1} (\sin 2\alpha) = 2\alpha = 2 \tan^{-1} x$.
Let $y = \tan \beta$. Then $\beta = \tan^{-1} y$.
$\cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) = \cos^{-1} \left( \frac{1-\tan^2 \beta}{1+\tan^2 \beta} \right) = \cos^{-1} (\cos 2\beta) = 2\beta = 2 \tan^{-1} y$.
Substituting these into the expression:
$\tan \frac{1}{2} [2 \tan^{-1} x + 2 \tan^{-1} y] = \tan [\tan^{-1} x + \tan^{-1} y]$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$:
$= \tan [\tan^{-1} \left( \frac{x+y}{1-xy} \right)] = \frac{x+y}{1-xy}$.

(A) Given the equation: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
We know that $\sin \left(\frac{\pi}{2}\right) = 1$. Therefore,we can write:
$\sin ^{-1} \frac{1}{5} + \cos ^{-1} x = \frac{\pi}{2}$
We also know the identity $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in [-1, 1]$.
Comparing $\sin ^{-1} \frac{1}{5} + \cos ^{-1} x = \sin ^{-1} \frac{1}{5} + \cos ^{-1} \frac{1}{5}$,we get:
$x = \frac{1}{5}$

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

(A) Let $\sin ^{-1} \frac{3}{5} = x$.
Then $\sin x = \frac{3}{5}$.
Since $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$,we have $\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}$.
Thus,$\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$.
Also,$\cot ^{-1} \frac{3}{2} = \tan ^{-1} \frac{2}{3}$.
Substituting these into the expression:
$\tan \left(\tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{2}{3}\right)$.
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left(\frac{a+b}{1-ab}\right)$:
$= \tan \left(\tan ^{-1} \left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \cdot \frac{2}{3}}\right)\right)$.
$= \tan \left(\tan ^{-1} \left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right)\right) = \tan \left(\tan ^{-1} \left(\frac{17/12}{6/12}\right)\right)$.
$= \tan \left(\tan ^{-1} \frac{17}{6}\right) = \frac{17}{6}$.

Let $\sin ^{-1} \frac{3}{5} = x$ and $\sin ^{-1} \frac{8}{17} = y$.
Then,$\sin x = \frac{3}{5}$ and $\sin y = \frac{8}{17}$.
Since $\cos x = \sqrt{1 - \sin^2 x}$,we have $\cos x = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Similarly,$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{1 - \frac{64}{289}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.
Using the identity $\cos(x - y) = \cos x \cos y + \sin x \sin y$,we get:
$\cos(x - y) = (\frac{4}{5})(\frac{15}{17}) + (\frac{3}{5})(\frac{8}{17}) = \frac{60}{85} + \frac{24}{85} = \frac{84}{85}$.
Thus,$x - y = \cos^{-1} \frac{84}{85}$.
Substituting back $x$ and $y$,we get $\sin^{-1} \frac{3}{5} - \sin^{-1} \frac{8}{17} = \cos^{-1} \frac{84}{85}$.

(C) Given expression: $\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]$
Divide the numerator and denominator by $b \cos x$:
$= \tan ^{-1}\left[\frac{\frac{a \cos x}{b \cos x} - \frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x} + \frac{a \sin x}{b \cos x}}\right]$
$= \tan ^{-1}\left[\frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b} \tan x}\right]$
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$,where $A = \frac{a}{b}$ and $B = \tan x$:
$= \tan ^{-1}\left(\frac{a}{b}\right) - \tan ^{-1}(\tan x)$
$= \tan ^{-1}\left(\frac{a}{b}\right) - x$

Let $\sin ^{-1} \frac{3}{5} = x$. Then $\sin x = \frac{3}{5}$.
Since $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Therefore,$\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}$.
This implies $x = \tan ^{-1} \frac{3}{4}$,so $\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{3}{4}$.
Now,consider the $L$.$H$.$S$.:
$2 \sin ^{-1} \frac{3}{5} = 2 \tan ^{-1} \frac{3}{4}$.
Using the formula $2 \tan ^{-1} A = \tan ^{-1} \left( \frac{2A}{1 - A^2} \right)$:
$2 \tan ^{-1} \frac{3}{4} = \tan ^{-1} \left( \frac{2 \times \frac{3}{4}}{1 - (\frac{3}{4})^2} \right)$.
$= \tan ^{-1} \left( \frac{3/2}{1 - 9/16} \right) = \tan ^{-1} \left( \frac{3/2}{7/16} \right)$.
$= \tan ^{-1} \left( \frac{3}{2} \times \frac{16}{7} \right) = \tan ^{-1} \frac{24}{7}$.
Thus,$L$.$H$.$S$. = $R$.$H$.$S$.

Let $\sin ^{-1} \frac{8}{17}=x$. Then,$\sin x=\frac{8}{17}$.
Since $\cos x=\sqrt{1-\sin^2 x}=\sqrt{1-(\frac{8}{17})^2}=\sqrt{\frac{289-64}{289}}=\frac{15}{17}$,we have $\tan x=\frac{\sin x}{\cos x}=\frac{8}{15}$.
Thus,$\sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15}$ $...(1)$
Let $\sin ^{-1} \frac{3}{5}=y$. Then,$\sin y=\frac{3}{5}$.
Since $\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-(\frac{3}{5})^2}=\sqrt{\frac{25-9}{25}}=\frac{4}{5}$,we have $\tan y=\frac{\sin y}{\cos y}=\frac{3}{4}$.
Thus,$\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$ $...(2)$
Now,taking the $L.H.S.$:
$\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4}$
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left(\frac{a+b}{1-ab}\right)$:
$= \tan ^{-1} \left(\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\right)$
$= \tan ^{-1} \left(\frac{\frac{32+45}{60}}{\frac{60-24}{60}}\right)$
$= \tan ^{-1} \left(\frac{77}{36}\right) = R.H.S.$

(A) Let $\cos ^{-1} \frac{4}{5} = x$. Then $\cos x = \frac{4}{5}$.
Since $\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{\frac{9}{25}} = \frac{3}{5}$,we have $\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{4/5} = \frac{3}{4}$.
Thus,$\cos ^{-1} \frac{4}{5} = \tan ^{-1} \frac{3}{4}$ $\dots (1)$.
Now,let $\cos ^{-1} \frac{12}{13} = y$. Then $\cos y = \frac{12}{13}$.
Since $\sin y = \sqrt{1 - (\frac{12}{13})^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$,we have $\tan y = \frac{5/13}{12/13} = \frac{5}{12}$.
Thus,$\cos ^{-1} \frac{12}{13} = \tan ^{-1} \frac{5}{12}$ $\dots (2)$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A + B}{1 - AB} \right)$:
$L.H.S = \tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{5}{12} = \tan ^{-1} \left( \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \times \frac{5}{12}} \right) = \tan ^{-1} \left( \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} \right) = \tan ^{-1} \left( \frac{56/48}{33/48} \right) = \tan ^{-1} \frac{56}{33}$.
Converting $\tan ^{-1} \frac{56}{33}$ back to $\cos ^{-1}$ form,let $\theta = \tan ^{-1} \frac{56}{33}$,then $\tan \theta = \frac{56}{33}$.
Using the triangle method,the hypotenuse is $\sqrt{56^2 + 33^2} = \sqrt{3136 + 1089} = \sqrt{4225} = 65$.
Thus,$\cos \theta = \frac{33}{65}$,which implies $\theta = \cos ^{-1} \frac{33}{65} = R.H.S$.