Properties of ITF Questions in English

(D) The general term of the series is $T_r = \cot^{-1}(r^2 + r + 1)$.
Using the identity $\cot^{-1} x = \tan^{-1} \left( \frac{1}{x} \right)$,we have $T_r = \tan^{-1} \left( \frac{1}{r^2 + r + 1} \right)$.
We can rewrite the denominator as $1 + r(r+1)$,so $T_r = \tan^{-1} \left( \frac{(r+1) - r}{1 + r(r+1)} \right)$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,we get $T_r = \tan^{-1}(r+1) - \tan^{-1}(r)$.
The sum of the first $n$ terms is $S_n = \sum_{r=1}^{n} (\tan^{-1}(r+1) - \tan^{-1}(r))$.
This is a telescoping series: $S_n = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1}(n+1) - \tan^{-1} n)$.
$S_n = \tan^{-1}(n+1) - \tan^{-1} 1$.
Using $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,$S_n = \tan^{-1} \left( \frac{n+1-1}{1+(n+1)(1)} \right) = \tan^{-1} \left( \frac{n}{n+2} \right)$.
Since $\tan^{-1} x = \cot^{-1} \left( \frac{1}{x} \right)$,$S_n = \cot^{-1} \left( \frac{n+2}{n} \right)$.
Thus,all options are correct.

(B) Let $x = \tan \theta$,then $\theta = \tan^{-1} x$.
Substituting this into the expression:
$\tan^{-1} \left( \frac{\sqrt{1 + \tan^2 \theta} - 1}{\tan \theta} \right) = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right)$
$= \tan^{-1} \left( \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} \right) = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right)$
Using the half-angle formulas $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$= \tan^{-1} \left( \frac{2 \sin^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \right) = \tan^{-1} \left( \tan \frac{\theta}{2} \right)$
$= \frac{\theta}{2} = \frac{1}{2} \tan^{-1} x$.

(A) We know that $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x)$.
Using the trigonometric identities $\sin \theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$ and $1 + \cos \theta = 2\cos^2(\frac{\theta}{2})$:
${\tan ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right] = {\tan ^{ - 1}}\left[ {\frac{{\sin (\pi /2 - x)}}{{1 + \cos (\pi /2 - x)}}} \right]$
$= {\tan ^{ - 1}}\left[ {\frac{{2\sin (\pi /4 - x/2)\cos (\pi /4 - x/2)}}{{2\cos^2 (\pi /4 - x/2)}}} \right]$
$= {\tan ^{ - 1}}\left[ {\tan (\pi /4 - x/2)} \right]$
$= \frac{\pi }{4} - \frac{x}{2}$.

(C) Let $x = \csc \theta$,where $\theta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.
Then $\theta = \csc^{-1} x$.
Substituting this into the expression:
$\tan^{-1} \left( \frac{1}{\sqrt{\csc^2 \theta - 1}} \right) = \tan^{-1} \left( \frac{1}{\sqrt{\cot^2 \theta}} \right)$
$= \tan^{-1} \left( \frac{1}{|\cot \theta|} \right)$.
Assuming $x > 1$,$\theta$ is in $(0, \frac{\pi}{2}]$,so $\cot \theta > 0$.
$= \tan^{-1} (\tan \theta) = \theta = \csc^{-1} x$.

(B) Let $x = \sin \theta$ and $\sqrt{x} = \sin \phi$.
Then,$\theta = \sin^{-1}(x)$ and $\phi = \sin^{-1}(\sqrt{x})$.
Substituting these into the expression:
${\sin ^{ - 1}}(x\sqrt {1 - x} - \sqrt x \,\sqrt {1 - {x^2}} )$
$= {\sin ^{ - 1}}(\sin \theta \sqrt {1 - {{\sin }^2}\phi } - \sin \phi \sqrt {1 - {{\sin }^2}\theta } )$
$= {\sin ^{ - 1}}(\sin \theta \cos \phi - \sin \phi \cos \theta )$
$= {\sin ^{ - 1}}(\sin (\theta - \phi ))$
$= \theta - \phi$
$= {\sin ^{ - 1}}(x) - {\sin ^{ - 1}}(\sqrt x )$.

(C) Given: ${\tan ^{ - 1}}\left( \frac{1 - x}{1 + x} \right) = \frac{1}{2}{\tan ^{ - 1}}x$
Let $x = \tan \theta$,then $\theta = {\tan ^{ - 1}}x$.
Substituting $x = \tan \theta$ in the equation:
${\tan ^{ - 1}}\left( \frac{1 - \tan \theta}{1 + \tan \theta} \right) = \frac{1}{2}\theta$
Using the formula $\tan \left( \frac{\pi}{4} - \theta \right) = \frac{\tan \frac{\pi}{4} - \tan \theta}{1 + \tan \frac{\pi}{4}\tan \theta} = \frac{1 - \tan \theta}{1 + \tan \theta}$:
${\tan ^{ - 1}}\left( \tan \left( \frac{\pi}{4} - \theta \right) \right) = \frac{\theta}{2}$
$\frac{\pi}{4} - \theta = \frac{\theta}{2}$
$\frac{\pi}{4} = \theta + \frac{\theta}{2} = \frac{3\theta}{2}$
$\theta = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$
Since $x = \tan \theta$,we have $x = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

(A) Let $\cos^{-1}x = \theta$,where $x \in [-1, 1]$.
Then $x = \cos \theta$,which implies $\cos \theta = x$.
We need to evaluate $\sin(\cot^{-1}(\tan \theta))$.
Since $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta} = \frac{\sqrt{1 - x^2}}{x}$.
Now,substitute this into the expression:
$\sin(\cot^{-1}(\frac{\sqrt{1 - x^2}}{x}))$.
Let $\cot^{-1}(\frac{\sqrt{1 - x^2}}{x}) = \phi$.
Then $\cot \phi = \frac{\sqrt{1 - x^2}}{x}$.
Using the identity $\csc^2 \phi = 1 + \cot^2 \phi = 1 + \frac{1 - x^2}{x^2} = \frac{x^2 + 1 - x^2}{x^2} = \frac{1}{x^2}$.
Thus,$\csc \phi = \frac{1}{x}$,which means $\sin \phi = x$.
Therefore,the value is $x$.

(C) Let $y = \sin^{-1} \sqrt{\frac{x}{x+a}}$.
Substitute $x = a \tan^2 \theta$,which implies $\tan \theta = \sqrt{\frac{x}{a}}$ or $\theta = \tan^{-1} \sqrt{\frac{x}{a}}$.
Then,$\sqrt{\frac{x}{x+a}} = \sqrt{\frac{a \tan^2 \theta}{a \tan^2 \theta + a}} = \sqrt{\frac{a \tan^2 \theta}{a(1 + \tan^2 \theta)}} = \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta$.
Thus,$y = \sin^{-1}(\sin \theta) = \theta$.
Substituting back,$y = \tan^{-1} \sqrt{\frac{x}{a}}$.

(D) Given the equation: $\sin \left( \sin^{-1} \frac{1}{5} + \cos^{-1} x \right) = 1$.
Taking $\sin^{-1}$ on both sides,we get: $\sin^{-1} \frac{1}{5} + \cos^{-1} x = \sin^{-1}(1)$.
Since $\sin^{-1}(1) = \frac{\pi}{2}$,the equation becomes: $\sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2}$.
We know the identity $\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2}$.
Comparing this with our equation,we have $\sin^{-1} \frac{1}{5} = \frac{\pi}{2} - \cos^{-1} x$.
Since $\frac{\pi}{2} - \cos^{-1} x = \sin^{-1} x$,we get $\sin^{-1} \frac{1}{5} = \sin^{-1} x$.
Therefore,$x = \frac{1}{5}$.

(B) Given that $\sin^{-1} x = \theta + \beta$ and $\sin^{-1} y = \theta - \beta$.
Therefore,$x = \sin(\theta + \beta)$ and $y = \sin(\theta - \beta)$.
Now,$1 + xy = 1 + \sin(\theta + \beta) \sin(\theta - \beta)$.
Using the identity $\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B$,we get:
$1 + xy = 1 + \sin^2 \theta - \sin^2 \beta$.
Since $1 - \sin^2 \beta = \cos^2 \beta$,we have:
$1 + xy = \sin^2 \theta + \cos^2 \beta$.

(C) We use the formula $\sin^{-1} a + \sin^{-1} b = \sin^{-1} \left( a \sqrt{1 - b^2} + b \sqrt{1 - a^2} \right)$.
Given $\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{2}{3} = \sin^{-1} x$.
Applying the formula with $a = \frac{1}{3}$ and $b = \frac{2}{3}$:
$\sin^{-1} x = \sin^{-1} \left( \frac{1}{3} \sqrt{1 - (\frac{2}{3})^2} + \frac{2}{3} \sqrt{1 - (\frac{1}{3})^2} \right)$
$= \sin^{-1} \left( \frac{1}{3} \sqrt{1 - \frac{4}{9}} + \frac{2}{3} \sqrt{1 - \frac{1}{9}} \right)$
$= \sin^{-1} \left( \frac{1}{3} \sqrt{\frac{5}{9}} + \frac{2}{3} \sqrt{\frac{8}{9}} \right)$
$= \sin^{-1} \left( \frac{1}{3} \cdot \frac{\sqrt{5}}{3} + \frac{2}{3} \cdot \frac{2\sqrt{2}}{3} \right)$
$= \sin^{-1} \left( \frac{\sqrt{5} + 4\sqrt{2}}{9} \right)$.
Therefore,$x = \frac{\sqrt{5} + 4\sqrt{2}}{9}$.

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

(B) Let $f(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right)$.
Using the property $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we can write:
$f(x) = \tan^{-1}(1) - \tan^{-1}(x) = \frac{\pi}{4} - \tan^{-1}(x)$.
Given the interval $0 \le x \le 1$,we know that $0 \le \tan^{-1}(x) \le \frac{\pi}{4}$.
Multiplying by $-1$,we get $-\frac{\pi}{4} \le -\tan^{-1}(x) \le 0$.
Adding $\frac{\pi}{4}$ to all parts,we get $\frac{\pi}{4} - \frac{\pi}{4} \le \frac{\pi}{4} - \tan^{-1}(x) \le \frac{\pi}{4} - 0$.
Thus,$0 \le f(x) \le \frac{\pi}{4}$.
The smallest value is $0$ and the largest value is $\frac{\pi}{4}$.

(B) Let $\sin^{-1} x = y$. Then $x = \sin y$.
Since $x$ is non-positive and permissible,we have $-1 \le x \le 0$.
This implies $-\frac{\pi}{2} \le \sin^{-1} x \le 0$,so $-\frac{\pi}{2} \le y \le 0$.
We know that $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ for $y \in [0, \pi]$.
Since $y \in [-\frac{\pi}{2}, 0]$,we have $-y \in [0, \frac{\pi}{2}]$.
Thus,$\cos(-y) = \sqrt{1 - x^2}$.
Since $\cos(-y) = \cos y$,we have $\cos y = \sqrt{1 - x^2}$.
Taking the inverse cosine on both sides for the range $[0, \frac{\pi}{2}]$,we get $-y = \cos^{-1} \sqrt{1 - x^2}$.
Therefore,$y = -\cos^{-1} \sqrt{1 - x^2}$.
Hence,$\sin^{-1} x = -\cos^{-1} \sqrt{1 - x^2}$.

(A) Let $\theta = \text{cosec}^{-1}x$. Then $\text{cosec}\theta = x$,where $|x| \ge 1$ and $\theta \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$.
We need to find $\sec(\text{cosec}^{-1}x) = \sec\theta$.
Since $\text{cosec}\theta = x$,we have $\sin\theta = \frac{1}{x}$.
Using the identity $\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$,we get $\cos\theta = \frac{\sqrt{x^2 - 1}}{|x|}$.
Therefore,$\sec\theta = \frac{1}{\cos\theta} = \frac{|x|}{\sqrt{x^2 - 1}}$.
Now,consider $\text{cosec}(\sec^{-1}x)$. Let $\phi = \sec^{-1}x$. Then $\sec\phi = x$,so $\cos\phi = \frac{1}{x}$.
Then $\sin\phi = \sqrt{1 - \cos^2\phi} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|}$.
Thus,$\text{cosec}\phi = \frac{1}{\sin\phi} = \frac{|x|}{\sqrt{x^2 - 1}}$.
Since both expressions equal $\frac{|x|}{\sqrt{x^2 - 1}}$,we have $\sec(\text{cosec}^{-1}x) = \text{cosec}(\sec^{-1}x)$.

(C) Given the equation: $\sin^{-1} x = 2\tan^{-1} x$
We know that $2\tan^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$ for $|x| \le 1$.
Substituting this into the equation,we get:
$\sin^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$
Taking the sine of both sides:
$x = \frac{2x}{1+x^2}$
Rearranging the terms:
$x(1+x^2) = 2x$
$x + x^3 = 2x$
$x^3 - x = 0$
$x(x^2 - 1) = 0$
$x(x-1)(x+1) = 0$
Thus,the solutions are $x = 0, 1, -1$.
Checking the domain: For all these values,$|x| \le 1$,so they are valid solutions.
The solution set is $\{-1, 0, 1\}$.

(C) We know that the property of inverse trigonometric functions states that $\tan^{-1}(\tan x) = x$ if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
Since $2$ radians is approximately $114.6^\circ$,which lies outside the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2}) \approx (-1.57, 1.57)$,we must adjust the expression.
However,the expression given is $\cos(\tan^{-1}(\tan 2))$.
Let $\theta = \tan^{-1}(\tan 2)$. Then $\tan \theta = \tan 2$.
This implies $\theta = 2 - \pi$ (since $2$ is in the second quadrant,$\tan 2 = \tan(2 - \pi)$ and $2 - \pi \approx -1.14$,which is in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$).
Thus,$\cos(\tan^{-1}(\tan 2)) = \cos(2 - \pi)$.
Using the identity $\cos(\theta - \pi) = \cos(\pi - \theta) = -\cos \theta$,we get $\cos(2 - \pi) = \cos(2)$.
Therefore,the value is $\cos 2$.

(B) Given that $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{\pi}{2}$.
Let $\sin^{-1} x = \alpha$,$\sin^{-1} y = \beta$,and $\sin^{-1} z = \gamma$.
Then $\alpha + \beta + \gamma = \frac{\pi}{2}$,which implies $\alpha + \beta = \frac{\pi}{2} - \gamma$.
Taking $\cos$ on both sides: $\cos(\alpha + \beta) = \cos(\frac{\pi}{2} - \gamma)$.
Using the identity $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ and $\cos(\frac{\pi}{2} - \gamma) = \sin \gamma$,we get:
$\cos \alpha \cos \beta - \sin \alpha \sin \beta = \sin \gamma$.
Since $\sin \alpha = x$,$\sin \beta = y$,and $\sin \gamma = z$,we have $\cos \alpha = \sqrt{1 - x^2}$ and $\cos \beta = \sqrt{1 - y^2}$.
Substituting these values: $\sqrt{1 - x^2} \sqrt{1 - y^2} - xy = z$.
Rearranging gives $\sqrt{1 - x^2} \sqrt{1 - y^2} = xy + z$.
Squaring both sides: $(1 - x^2)(1 - y^2) = (xy + z)^2$.
$1 - x^2 - y^2 + x^2 y^2 = x^2 y^2 + 2xyz + z^2$.
$1 = x^2 + y^2 + z^2 + 2xyz$.

(C) We know that $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,which implies $\frac{\pi}{2} - \sin^{-1}(x) = \cos^{-1}(x)$.
Substituting $x = -\frac{\sqrt{3}}{2}$,we get:
$\sin \left[ \cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) \right]$
Since $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,we have:
$\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right) = \pi - \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Now,$\sin \left( \frac{5\pi}{6} \right) = \sin \left( \pi - \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}$.

(D) Let $\tan^{-1}x = \theta$,then $\tan \theta = x$.
We know that $\cos(\tan^{-1}x) = \cos \theta = \frac{1}{\sqrt{1+x^2}}$.
Now,the expression becomes $\sin[\cot^{-1}(\frac{1}{\sqrt{1+x^2}})]$.
Let $\cot^{-1}(\frac{1}{\sqrt{1+x^2}}) = \phi$,then $\cot \phi = \frac{1}{\sqrt{1+x^2}}$.
This implies $\tan \phi = \sqrt{1+x^2}$.
Using the identity $\sin \phi = \frac{\tan \phi}{\sqrt{1+\tan^2 \phi}}$,we get:
$\sin \phi = \frac{\sqrt{1+x^2}}{\sqrt{1+(1+x^2)}} = \sqrt{\frac{1+x^2}{2+x^2}}$.
Thus,the correct option is $D$.

(A) Let $x = \cos ^{ - 1}\frac{4}{5}$. Then $\cos x = \frac{4}{5}$.
Since $\cos x = \frac{\text{adj}}{\text{hyp}} = \frac{4}{5}$,the opposite side is $\sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3$.
Thus,$\tan x = \frac{\text{opp}}{\text{adj}} = \frac{3}{4}$,which implies $x = \tan ^{ - 1}\frac{3}{4}$.
Now,the expression becomes $\tan ^{ - 1}\frac{3}{4} + \tan ^{ - 1}\frac{3}{5}$.
Using the formula $\tan ^{ - 1}A + \tan ^{ - 1}B = \tan ^{ - 1}\left( \frac{A + B}{1 - AB} \right)$:
$= \tan ^{ - 1}\left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4} \times \frac{3}{5}} \right)$
$= \tan ^{ - 1}\left( \frac{\frac{15 + 12}{20}}{1 - \frac{9}{20}} \right)$
$= \tan ^{ - 1}\left( \frac{\frac{27}{20}}{\frac{11}{20}} \right)$
$= \tan ^{ - 1}\frac{27}{11}$.

(A) We know that $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$.
Given expression: $\sin^{-1}x + \sin^{-1}\frac{1}{x} + \cos^{-1}x + \cos^{-1}\frac{1}{x}$.
Rearranging the terms,we get:
$= (\sin^{-1}x + \cos^{-1}x) + (\sin^{-1}\frac{1}{x} + \cos^{-1}\frac{1}{x})$.
Using the identity $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$,we have:
$= \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
Thus,the correct option is $A$.

(D) We use the formula $2\tan ^{-1}x = \tan ^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$.
Applying this to $2\tan ^{-1}\frac{1}{3}$:
$2\tan ^{-1}\frac{1}{3} = \tan ^{-1}\left(\frac{2(1/3)}{1-(1/3)^2}\right) = \tan ^{-1}\left(\frac{2/3}{1-1/9}\right) = \tan ^{-1}\left(\frac{2/3}{8/9}\right) = \tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$.
Now,the expression becomes $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{1}{2}\right)$.
Using the formula $\tan ^{-1}x + \tan ^{-1}y = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$:
$\tan ^{-1}\left(\frac{3/4 + 1/2}{1 - (3/4)(1/2)}\right) = \tan ^{-1}\left(\frac{5/4}{1 - 3/8}\right) = \tan ^{-1}\left(\frac{5/4}{5/8}\right) = \tan ^{-1}\left(\frac{5}{4} \times \frac{8}{5}\right) = \tan ^{-1}(2)$.
Thus,the correct option is $D$.

(B) Let $\theta = \cos^{-1} \frac{4}{5}$. Then $\cos \theta = \frac{4}{5}$.
Since $\tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta} = \frac{\sqrt{1 - (16/25)}}{4/5} = \frac{3/5}{4/5} = \frac{3}{4}$,we have $\theta = \tan^{-1} \frac{3}{4}$.
Substituting this into the expression,we get $\tan \left[ \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} \right]$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$,we have:
$\tan \left[ \tan^{-1} \left( \frac{3/4 + 2/3}{1 - (3/4)(2/3)} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{9/12 + 8/12}{1 - 6/12} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{17/12}{6/12} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{17}{6} \right) \right] = \frac{17}{6}$.

(D) We use the formula for ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y$ when $xy > 1$:
${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi + {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$.
First,calculate ${\tan ^{ - 1}}2 + {\tan ^{ - 1}}3$:
Since $2 \times 3 = 6 > 1$,we have:
${\tan ^{ - 1}}2 + {\tan ^{ - 1}}3 = \pi + {\tan ^{ - 1}}\left( {\frac{{2 + 3}}{{1 - 2 \times 3}}} \right) = \pi + {\tan ^{ - 1}}\left( {\frac{5}{{1 - 6}}} \right) = \pi + {\tan ^{ - 1}}\left( {\frac{5}{{ - 5}}} \right) = \pi + {\tan ^{ - 1}}( - 1) = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$.
Now,add ${\tan ^{ - 1}}1$:
${\tan ^{ - 1}}1 + (\frac{{3\pi }}{4}) = \frac{\pi }{4} + \frac{{3\pi }}{4} = \frac{{4\pi }}{4} = \pi$.
Since $\pi$ is not among the options $A, B, C$,the correct answer is $D$.

(A) Let $\cot^{-1} \frac{3}{4} = \theta$,then $\cot \theta = \frac{3}{4}$.
Since $\cot \theta = \frac{\text{base}}{\text{perpendicular}} = \frac{3}{4}$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.
Therefore,$\sin \theta = \frac{4}{5}$,which implies $\theta = \sin^{-1} \frac{4}{5}$.
Now,the expression becomes $\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13}$.
Using the formula $\sin^{-1} x + \sin^{-1} y = \sin^{-1} \left( x \sqrt{1 - y^2} + y \sqrt{1 - x^2} \right)$:
$= \sin^{-1} \left( \frac{4}{5} \sqrt{1 - \left( \frac{5}{13} \right)^2} + \frac{5}{13} \sqrt{1 - \left( \frac{4}{5} \right)^2} \right)$
$= \sin^{-1} \left( \frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5} \right)$
$= \sin^{-1} \left( \frac{48}{65} + \frac{15}{65} \right)$
$= \sin^{-1} \frac{63}{65}$.

(D) Given that $\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = \pi$.
$\Rightarrow \cos^{-1} x + \cos^{-1} y = \pi - \cos^{-1} z$.
Using the identity $\pi - \cos^{-1} z = \cos^{-1}(-z)$,we get:
$\cos^{-1} x + \cos^{-1} y = \cos^{-1}(-z)$.
Applying the formula $\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$:
$\cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2}) = \cos^{-1}(-z)$.
Taking cosine on both sides:
$xy - \sqrt{1-x^2}\sqrt{1-y^2} = -z$.
Rearranging terms:
$xy + z = \sqrt{1-x^2}\sqrt{1-y^2}$.
Squaring both sides:
$(xy + z)^2 = (1-x^2)(1-y^2)$.
$x^2y^2 + z^2 + 2xyz = 1 - x^2 - y^2 + x^2y^2$.
$x^2 + y^2 + z^2 + 2xyz = 1$.
Alternatively,let $x = y = z = \frac{1}{2}$. Then $\cos^{-1}(\frac{1}{2}) + \cos^{-1}(\frac{1}{2}) + \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} + \frac{\pi}{3} + \frac{\pi}{3} = \pi$. Substituting these values into option $(d)$: $(\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 + 2(\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{2}{8} = \frac{3}{4} + \frac{1}{4} = 1$. Thus,option $(d)$ is correct.

(C) We are given the equation: ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}A$.
Using the standard trigonometric identity for the difference of inverse tangents:
${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( \frac{x - y}{1 + xy} \right)$.
Comparing this with the given equation:
${\tan ^{ - 1}}\left( \frac{x - y}{1 + xy} \right) = {\tan ^{ - 1}}A$.
Therefore,by equating the arguments,we get:
$A = \frac{x - y}{1 + xy}$.

(D) Given that $\tan ^{-1}x + \tan ^{-1}y + \tan ^{-1}z = \frac{\pi }{2}.$
Using the formula $\tan ^{-1}x + \tan ^{-1}y + \tan ^{-1}z = \tan ^{-1}\left( \frac{x + y + z - xyz}{1 - xy - yz - zx} \right),$
$\tan ^{-1}\left( \frac{x + y + z - xyz}{1 - xy - yz - zx} \right) = \frac{\pi }{2}.$
Taking tangent on both sides:
$\frac{x + y + z - xyz}{1 - xy - yz - zx} = \tan \left( \frac{\pi }{2} \right) = \infty.$
For the expression to be undefined (approaching $\infty$),the denominator must be zero:
$1 - xy - yz - zx = 0.$
Rearranging the terms,we get:
$xy + yz + zx = 1$ or $xy + yz + zx - 1 = 0.$
Thus,option $(d)$ is correct.

(C) Given the equation: ${\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}$
Using the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( \frac{A+B}{1-AB} \right)$,we get:
${\tan ^{ - 1}}\left[ \frac{{\frac{{x - 1}}{{x + 2}} + \frac{{x + 1}}{{x + 2}}}}{{1 - \left( {\frac{{x - 1}}{{x + 2}}} \right)\left( {\frac{{x + 1}}{{x + 2}}} \right)}} \right] = \frac{\pi }{4}$
Simplifying the numerator and denominator:
${\tan ^{ - 1}}\left[ \frac{{\frac{{2x}}{{x + 2}}}}{{\frac{{{{(x + 2)}^2} - ({x^2} - 1)}}{{{{(x + 2)}^2}}}}} \right] = \frac{\pi }{4}$
${\tan ^{ - 1}}\left[ \frac{{2x(x + 2)}}{{{x^2} + 4x + 4 - {x^2} + 1}} \right] = \frac{\pi }{4}$
${\tan ^{ - 1}}\left[ \frac{{2x(x + 2)}}{{4x + 5}} \right] = \frac{\pi }{4}$
Taking tangent on both sides:
$\frac{{2{x^2} + 4x}}{{4x + 5}} = \tan \frac{\pi }{4} = 1$
$2{x^2} + 4x = 4x + 5$
$2{x^2} = 5$
${x^2} = \frac{5}{2}$
$x = \pm \sqrt{\frac{5}{2}}$

(B) We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Given expression: $\cos \left[ \cos^{-1}\frac{1}{5} + \cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5} \right]$.
Using the property,this simplifies to: $\cos \left[ \frac{\pi}{2} + \cos^{-1}\frac{1}{5} \right]$.
Since $\cos(\frac{\pi}{2} + \theta) = -\sin\theta$,we have: $-\sin \left( \cos^{-1}\frac{1}{5} \right)$.
Let $\cos^{-1}\frac{1}{5} = \alpha$,then $\cos\alpha = \frac{1}{5}$.
Then $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (\frac{1}{5})^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$.
Thus,the expression equals $-\frac{2\sqrt{6}}{5}$.

(B) We use the standard identity for the difference of inverse tangents: $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$.
Applying this to the given expression:
$\tan^{-1} \left( \frac{a - b}{1 + ab} \right) = \tan^{-1} a - \tan^{-1} b$
Similarly:
$\tan^{-1} \left( \frac{b - c}{1 + bc} \right) = \tan^{-1} b - \tan^{-1} c$
Adding these two results together:
$(\tan^{-1} a - \tan^{-1} b) + (\tan^{-1} b - \tan^{-1} c)$
The term $-\tan^{-1} b$ and $+\tan^{-1} b$ cancel each other out:
$= \tan^{-1} a - \tan^{-1} c$.

(B) Given 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:
$\tan^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} \right) = \frac{\pi}{4}$
$\tan^{-1} \left( \frac{5x}{1 - 6x^2} \right) = \tan^{-1}(1)$
$\frac{5x}{1 - 6x^2} = 1$
$1 - 6x^2 = 5x$
$6x^2 + 5x - 1 = 0$
$(6x - 1)(x + 1) = 0$
So,$x = \frac{1}{6}$ or $x = -1$.
If $x = -1$,then $\tan^{-1}(-2) + \tan^{-1}(-3) = -(\tan^{-1} 2 + \tan^{-1} 3)$,which is negative and cannot be $\frac{\pi}{4}$.
If $x = \frac{1}{6}$,then $\tan^{-1}(\frac{1}{3}) + \tan^{-1}(\frac{1}{2}) = \tan^{-1} \left( \frac{1/3 + 1/2}{1 - 1/6} \right) = \tan^{-1} \left( \frac{5/6}{5/6} \right) = \tan^{-1}(1) = \frac{\pi}{4}$.
Thus,the only valid solution is $x = \frac{1}{6}$.

(A) We know that $2{\sin ^{ - 1}}x = {\sin ^{ - 1}}(2x\sqrt {1 - {x^2}} )$ for $|x| \le \frac{1}{\sqrt{2}}$.
Here,$x = \frac{3}{5}$,so $2{\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\left( 2 \times \frac{3}{5} \times \sqrt {1 - \frac{9}{{25}}} \right) = {\sin ^{ - 1}}\left( \frac{6}{5} \times \frac{4}{5} \right) = {\sin ^{ - 1}}\frac{{24}}{{25}}$.
Now,the expression becomes ${\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}}$.
Using the identity ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$,we get ${\sin ^{ - 1}}\frac{{24}}{{25}} + {\cos ^{ - 1}}\frac{{24}}{{25}} = \frac{\pi }{2}$.

(A) We use the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$ for $xy < 1$.
Here,$x = \frac{1}{3}$ and $y = \frac{1}{2}$.
Since $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6} < 1$,the formula is applicable.
$\tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{2} = \tan^{-1} \left( \frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3} \times \frac{1}{2}} \right) = \tan^{-1} \left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} \right) = \tan^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) = \tan^{-1}(1) = \frac{\pi}{4}$.
Therefore,$\cos \left[ \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{2} \right] = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$.

(D) We know that ${\cot ^{ - 1}}(y) = {\tan ^{ - 1}}\left( \frac{1}{y} \right)$ for $y > 0$.
Given expression is ${\tan ^{ - 1}}x + {\cot ^{ - 1}}(x + 1)$.
Using the property,we can write this as:
${\tan ^{ - 1}}x + {\tan ^{ - 1}}\left( \frac{1}{x + 1} \right)$.
Now,apply the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( \frac{A + B}{1 - AB} \right)$:
$= {\tan ^{ - 1}}\left( \frac{x + \frac{1}{x + 1}}{1 - x \cdot \frac{1}{x + 1}} \right)$
$= {\tan ^{ - 1}}\left( \frac{\frac{x(x + 1) + 1}{x + 1}}{\frac{x + 1 - x}{x + 1}} \right)$
$= {\tan ^{ - 1}}\left( \frac{x^2 + x + 1}{1} \right)$
$= {\tan ^{ - 1}}({x^2} + x + 1)$.

(A) We know the identity for the difference of inverse cotangent functions: $\cot ^{ - 1}\left(\frac{ab + 1}{a - b}\right) = \cot ^{ - 1}b - \cot ^{ - 1}a$.
Applying this identity to each term in the expression:
$\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) = \cot ^{ - 1}y - \cot ^{ - 1}x$
$\cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) = \cot ^{ - 1}z - \cot ^{ - 1}y$
$\cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right) = \cot ^{ - 1}x - \cot ^{ - 1}z$
Adding these three expressions together:
$(\cot ^{ - 1}y - \cot ^{ - 1}x) + (\cot ^{ - 1}z - \cot ^{ - 1}y) + (\cot ^{ - 1}x - \cot ^{ - 1}z)$
Canceling the terms:
$= (\cot ^{ - 1}y - \cot ^{ - 1}y) + (\cot ^{ - 1}z - \cot ^{ - 1}z) + (\cot ^{ - 1}x - \cot ^{ - 1}x) = 0$
Thus,the correct option is $A$.

(C) Given equation: $\tan^{-1} \left( \frac{a+x}{a} \right) + \tan^{-1} \left( \frac{a-x}{a} \right) = \frac{\pi}{6}$
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$:
$\tan^{-1} \left( \frac{\frac{a+x}{a} + \frac{a-x}{a}}{1 - \left( \frac{a+x}{a} \right) \left( \frac{a-x}{a} \right)} \right) = \frac{\pi}{6}$
$\tan^{-1} \left( \frac{\frac{2a}{a}}{1 - \frac{a^2-x^2}{a^2}} \right) = \frac{\pi}{6}$
$\tan^{-1} \left( \frac{2}{\frac{a^2 - a^2 + x^2}{a^2}} \right) = \frac{\pi}{6}$
$\tan^{-1} \left( \frac{2a^2}{x^2} \right) = \frac{\pi}{6}$
$\frac{2a^2}{x^2} = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$
$x^2 = 2\sqrt{3} a^2$

(B) Given equation: $\cos^{-1} \frac{3}{5} - \sin^{-1} \frac{4}{5} = \cos^{-1} x$
We know that $\sin^{-1} \theta = \cos^{-1} \sqrt{1 - \theta^2}$ for $\theta \in [0, 1]$.
Therefore,$\sin^{-1} \frac{4}{5} = \cos^{-1} \sqrt{1 - (\frac{4}{5})^2} = \cos^{-1} \sqrt{1 - \frac{16}{25}} = \cos^{-1} \sqrt{\frac{9}{25}} = \cos^{-1} \frac{3}{5}$.
Substituting this back into the original equation:
$\cos^{-1} \frac{3}{5} - \cos^{-1} \frac{3}{5} = \cos^{-1} x$
$0 = \cos^{-1} x$
$x = \cos(0) = 1$.

(B) We know that ${\csc ^{ - 1}}x = {\cot ^{ - 1}}\sqrt {{x^2} - 1} $ for $x > 1$.
Substituting $x = \sqrt 5$,we get ${\csc ^{ - 1}}\sqrt 5 = {\cot ^{ - 1}}\sqrt {{(\sqrt 5 )^2} - 1} = {\cot ^{ - 1}}\sqrt {5 - 1} = {\cot ^{ - 1}}\sqrt 4 = {\cot ^{ - 1}}2$.
Now,the expression becomes ${\cot ^{ - 1}}3 + {\cot ^{ - 1}}2$.
Using the formula ${\cot ^{ - 1}}x + {\cot ^{ - 1}}y = {\cot ^{ - 1}}\left( {\frac{{xy - 1}}{{x + y}}} \right)$ for $xy > 1$:
${\cot ^{ - 1}}3 + {\cot ^{ - 1}}2 = {\cot ^{ - 1}}\left( {\frac{{3 \times 2 - 1}}{{3 + 2}}} \right) = {\cot ^{ - 1}}\left( {\frac{{6 - 1}}{5}} \right) = {\cot ^{ - 1}}\left( {\frac{5}{5}} \right) = {\cot ^{ - 1}}(1)$.
Since ${\cot ^{ - 1}}(1) = \frac{\pi }{4}$,the final answer is $\frac{\pi }{4}$.

(B) Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
The given expression is $\tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$.
Substitute $x = \tan \theta$:
$= \tan^{-1} \left( \frac{1 - \tan^2 \theta}{2 \tan \theta} \right) + \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right)$
Using trigonometric identities $\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}$ and $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$:
$= \tan^{-1} (\cot 2\theta) + \cos^{-1} (\cos 2\theta)$
Since $\tan^{-1} (\cot 2\theta) = \tan^{-1} \left( \tan \left( \frac{\pi}{2} - 2\theta \right) \right) = \frac{\pi}{2} - 2\theta$ and $\cos^{-1} (\cos 2\theta) = 2\theta$:
$= \left( \frac{\pi}{2} - 2\theta \right) + 2\theta$
$= \frac{\pi}{2}$.

(D) Given equation: ${\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x$
Rearranging the terms: ${\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x - {\tan ^{ - 1}}x$
Applying the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( \frac{A+B}{1-AB} \right)$ and ${\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( \frac{A-B}{1+AB} \right)$:
${\tan ^{ - 1}}\left( \frac{(x-1)+(x+1)}{1-(x-1)(x+1)} \right) = {\tan ^{ - 1}}\left( \frac{3x-x}{1+(3x)(x)} \right)$
${\tan ^{ - 1}}\left( \frac{2x}{1-(x^2-1)} \right) = {\tan ^{ - 1}}\left( \frac{2x}{1+3x^2} \right)$
This implies: $\frac{2x}{2-x^2} = \frac{2x}{1+3x^2}$
Case $1$: $2x = 0 \Rightarrow x = 0$
Case $2$: $\frac{1}{2-x^2} = \frac{1}{1+3x^2}$
$1 + 3x^2 = 2 - x^2$
$4x^2 = 1 \Rightarrow x^2 = \frac{1}{4} \Rightarrow x = \pm \frac{1}{2}$
Thus,the solutions are $x = 0, \pm \frac{1}{2}$.

(B) We know that for any $x, y \in [-1, 1]$,the range of $\cos^{-1} x$ is $[0, \pi]$.
Given $\cos^{-1} x + \cos^{-1} y = 2\pi$.
Since the maximum value of $\cos^{-1} x$ is $\pi$,the only way their sum can be $2\pi$ is if $\cos^{-1} x = \pi$ and $\cos^{-1} y = \pi$.
This implies $x = \cos(\pi) = -1$ and $y = \cos(\pi) = -1$.
Now,we need to find $\sin^{-1} x + \sin^{-1} y$.
Substituting the values of $x$ and $y$:
$\sin^{-1}(-1) + \sin^{-1}(-1) = -\frac{\pi}{2} - \frac{\pi}{2} = -\pi$.
Alternatively,using the identity $\cos^{-1} t = \frac{\pi}{2} - \sin^{-1} t$:
$(\frac{\pi}{2} - \sin^{-1} x) + (\frac{\pi}{2} - \sin^{-1} y) = 2\pi$.
$\pi - (\sin^{-1} x + \sin^{-1} y) = 2\pi$.
$\sin^{-1} x + \sin^{-1} y = \pi - 2\pi = -\pi$.

(B) Let $\alpha = \sin^{-1} \frac{1}{\sqrt{5}}$. Then $\sin \alpha = \frac{1}{\sqrt{5}}$.
Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$,we have $\cos \alpha = \sqrt{1 - \frac{1}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.
Thus,$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$.
Therefore,$\alpha = \tan^{-1} \frac{1}{2} = \cot^{-1} 2$.
Now,the expression becomes $\cot^{-1} 2 + \cot^{-1} 3$.
Using the formula $\cot^{-1} x + \cot^{-1} y = \cot^{-1} \left( \frac{xy - 1}{x + y} \right)$ for $x, y > 0$:
$\cot^{-1} 2 + \cot^{-1} 3 = \cot^{-1} \left( \frac{2 \times 3 - 1}{2 + 3} \right) = \cot^{-1} \left( \frac{5}{5} \right) = \cot^{-1} (1) = \frac{\pi}{4}$.

(D) We are given the equation $\cot^{-1} \alpha + \cot^{-1} \beta = \cot^{-1} x$.
Using the standard identity for the sum of inverse cotangent functions,$\cot^{-1} \alpha + \cot^{-1} \beta = \cot^{-1} \left( \frac{\alpha \beta - 1}{\alpha + \beta} \right)$.
Comparing this with the given equation $\cot^{-1} x$,we get $x = \frac{\alpha \beta - 1}{\alpha + \beta}$.
Thus,the correct option is $D$.

(D) Given the equation: ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x$
Substitute $a = \tan \theta$ and $b = \tan \phi$:
${\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + {\sin ^{ - 1}}\left( {\frac{{2\tan \phi }}{{1 + {{\tan }^2}\phi }}} \right) = 2{\tan ^{ - 1}}x$
Using the identity $\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}$,we get:
${\sin ^{ - 1}}(\sin 2\theta) + {\sin ^{ - 1}}(\sin 2\phi) = 2{\tan ^{ - 1}}x$
Simplifying the expression:
$2\theta + 2\phi = 2{\tan ^{ - 1}}x$
Dividing by $2$:
$\theta + \phi = {\tan ^{ - 1}}x$
Taking the tangent of both sides:
$x = \tan(\theta + \phi)$
Using the formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:
$x = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}$
Substituting $a$ and $b$ back:
$x = \frac{a + b}{1 - ab}$

(A) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ for $xy < 1$.
First,evaluate $\tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{3}{5}$:
$\tan ^{-1} \left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4} \times \frac{3}{5}} \right) = \tan ^{-1} \left( \frac{\frac{15+12}{20}}{1 - \frac{9}{20}} \right) = \tan ^{-1} \left( \frac{27/20}{11/20} \right) = \tan ^{-1} \frac{27}{11}$.
Now,subtract $\tan ^{-1} \frac{8}{19}$ using $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \left( \frac{x-y}{1+xy} \right)$:
$\tan ^{-1} \frac{27}{11} - \tan ^{-1} \frac{8}{19} = \tan ^{-1} \left( \frac{\frac{27}{11} - \frac{8}{19}}{1 + \frac{27}{11} \times \frac{8}{19}} \right)$.
Calculate the numerator: $\frac{27 \times 19 - 8 \times 11}{11 \times 19} = \frac{513 - 88}{209} = \frac{425}{209}$.
Calculate the denominator: $1 + \frac{216}{209} = \frac{209 + 216}{209} = \frac{425}{209}$.
Thus,$\tan ^{-1} \left( \frac{425/209}{425/209} \right) = \tan ^{-1}(1) = \frac{\pi }{4}$.

(C) We use the formula $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$.
First,$4 \tan^{-1} \frac{1}{5} = 2 \left( 2 \tan^{-1} \frac{1}{5} \right) = 2 \tan^{-1} \left( \frac{2/5}{1 - 1/25} \right) = 2 \tan^{-1} \left( \frac{2/5}{24/25} \right) = 2 \tan^{-1} \left( \frac{5}{12} \right)$.
Now,$2 \tan^{-1} \frac{5}{12} = \tan^{-1} \left( \frac{2(5/12)}{1 - (5/12)^2} \right) = \tan^{-1} \left( \frac{5/6}{1 - 25/144} \right) = \tan^{-1} \left( \frac{5/6}{119/144} \right) = \tan^{-1} \left( \frac{5}{6} \times \frac{144}{119} \right) = \tan^{-1} \left( \frac{120}{119} \right)$.
Next,we evaluate $\tan^{-1} \frac{1}{99} - \tan^{-1} \frac{1}{70} = \tan^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} \right) = \tan^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} \right) = \tan^{-1} \left( \frac{-29}{6931} \right) = -\tan^{-1} \left( \frac{1}{239} \right)$.
Finally,$\tan^{-1} \frac{120}{119} - \tan^{-1} \frac{1}{239} = \tan^{-1} \left( \frac{120/119 - 1/239}{1 + (120/119)(1/239)} \right) = \tan^{-1} \left( \frac{(28680 - 119) / 28441}{(28441 + 120) / 28441} \right) = \tan^{-1} \left( \frac{28561}{28561} \right) = \tan^{-1}(1) = \frac{\pi}{4}$.

(B) We know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Therefore,$\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$ and $\sin^{-1} y = \frac{\pi}{2} - \cos^{-1} y$.
Substituting these into the given equation:
$(\frac{\pi}{2} - \cos^{-1} x) + (\frac{\pi}{2} - \cos^{-1} y) = \frac{2\pi}{3}$
$\pi - (\cos^{-1} x + \cos^{-1} y) = \frac{2\pi}{3}$
$\cos^{-1} x + \cos^{-1} y = \pi - \frac{2\pi}{3}$
$\cos^{-1} x + \cos^{-1} y = \frac{\pi}{3}$.

(A) Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$,we have:
$\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \times \frac{2}{9}}\right)$
$= \tan^{-1}\left(\frac{\frac{9+8}{36}}{1 - \frac{2}{36}}\right) = \tan^{-1}\left(\frac{17/36}{34/36}\right) = \tan^{-1}\left(\frac{17}{34}\right) = \tan^{-1}\left(\frac{1}{2}\right)$
Now,we use the formula $2\tan^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$:
$\tan^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} \left( 2\tan^{-1}\left(\frac{1}{2}\right) \right) = \frac{1}{2} \cos^{-1}\left(\frac{1-(1/2)^2}{1+(1/2)^2}\right)$
$= \frac{1}{2} \cos^{-1}\left(\frac{1-1/4}{1+1/4}\right) = \frac{1}{2} \cos^{-1}\left(\frac{3/4}{5/4}\right) = \frac{1}{2} \cos^{-1}\left(\frac{3}{5}\right)$
Thus,the correct option is $A$.