Properties of ITF Questions in English

(A) Given expression: ${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right)$
Divide the numerator and denominator of the second term by $x$:
${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{1 - y/x}}{{1 + y/x}}} \right)$
Using the formula ${\tan ^{ - 1}}A - {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\frac{{A - B}}{{1 + AB}}} \right)$,we can write:
${\tan ^{ - 1}}\left( {\frac{{1 - y/x}}{{1 + 1 \cdot (y/x)}}} \right) = {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$
Substituting this back into the expression:
${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - \left( {\frac{\pi }{4} - {\tan ^{ - 1}}\left( {\frac{y}{x}} \right)} \right)$
Since ${\tan ^{ - 1}}\left( {\frac{y}{x}} \right) = {\cot ^{ - 1}}\left( {\frac{x}{y}} \right)$:
${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) + {\cot ^{ - 1}}\left( {\frac{x}{y}} \right) - \frac{\pi }{4}$
Using the identity ${\tan ^{ - 1}}\theta + {\cot ^{ - 1}}\theta = \frac{\pi }{2}$:
$\frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}$.

(D) We use the formula $2{\tan ^{ - 1}}(x) = {\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)$ for $|x| < 1$.
First,simplify $2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right)$:
$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\frac{{2(1/3)}}{{1 - {{(1/3)}^2}}}} \right) = {\tan ^{ - 1}}\left( {\frac{{2/3}}{{1 - 1/9}}} \right) = {\tan ^{ - 1}}\left( {\frac{{2/3}}{{8/9}}} \right) = {\tan ^{ - 1}}\left( {\frac{2}{3} \times \frac{9}{8}} \right) = {\tan ^{ - 1}}\left( {\frac{3}{4}} \right)$.
Now,add the remaining term using the formula ${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y) = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)$:
${\tan ^{ - 1}}\left( {\frac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = {\tan ^{ - 1}}\left( {\frac{{3/4 + 1/7}}{{1 - (3/4)(1/7)}}} \right)$.
Calculate the numerator and denominator:
Numerator: $\frac{3}{4} + \frac{1}{7} = \frac{{21 + 4}}{{28}} = \frac{{25}}{{28}}$.
Denominator: $1 - \frac{3}{{28}} = \frac{{28 - 3}}{{28}} = \frac{{25}}{{28}}$.
Thus,${\tan ^{ - 1}}\left( {\frac{{25/28}}{{25/28}}} \right) = {\tan ^{ - 1}}(1) = \frac{\pi }{4}$.

(D) We are given the expression: $\cos^{-1}\left(\frac{15}{17}\right) + 2\tan^{-1}\left(\frac{1}{5}\right)$.
First,convert $2\tan^{-1}\left(\frac{1}{5}\right)$ into $\cos^{-1}$ form using the formula $2\tan^{-1}(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$:
$2\tan^{-1}\left(\frac{1}{5}\right) = \cos^{-1}\left(\frac{1 - (1/5)^2}{1 + (1/5)^2}\right) = \cos^{-1}\left(\frac{1 - 1/25}{1 + 1/25}\right) = \cos^{-1}\left(\frac{24/25}{26/25}\right) = \cos^{-1}\left(\frac{12}{13}\right)$.
Now,the expression becomes $\cos^{-1}\left(\frac{15}{17}\right) + \cos^{-1}\left(\frac{12}{13}\right)$.
Using the formula $\cos^{-1}(x) + \cos^{-1}(y) = \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}\right)$:
$\cos^{-1}\left(\frac{15}{17} \times \frac{12}{13} - \sqrt{1 - \left(\frac{15}{17}\right)^2} \sqrt{1 - \left(\frac{12}{13}\right)^2}\right)$
$= \cos^{-1}\left(\frac{180}{221} - \sqrt{1 - \frac{225}{289}} \sqrt{1 - \frac{144}{169}}\right)$
$= \cos^{-1}\left(\frac{180}{221} - \sqrt{\frac{64}{289}} \sqrt{\frac{25}{169}}\right)$
$= \cos^{-1}\left(\frac{180}{221} - \frac{8}{17} \times \frac{5}{13}\right)$
$= \cos^{-1}\left(\frac{180}{221} - \frac{40}{221}\right) = \cos^{-1}\left(\frac{140}{221}\right)$.
Since $\cos^{-1}\left(\frac{140}{221}\right)$ is not among the options,the correct choice is $(d)$.

(A) Let $\theta = \sin ^{-1}\left(\frac{3}{5}\right)$. Then $\sin \theta = \frac{3}{5}$.
Using the identity $\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin ^2 \theta}} = \frac{3/5}{\sqrt{1 - (3/5)^2}} = \frac{3/5}{4/5} = \frac{3}{4}$.
Thus,$\sin ^{-1}\left(\frac{3}{5}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$.
Now,the expression becomes $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{1}{7}\right)$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy < 1$:
$\tan ^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \times \frac{1}{7}}\right) = \tan ^{-1}\left(\frac{\frac{21+4}{28}}{1 - \frac{3}{28}}\right) = \tan ^{-1}\left(\frac{25/28}{25/28}\right) = \tan ^{-1}(1) = \frac{\pi }{4}$.

(C) Given equation: $\tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2}$
We know that $\tan^{-1}(A) + \tan^{-1}(B) = \frac{\pi}{2}$ implies $AB = 1$ (for $A, B > 0$).
Here,$A = 1 + x$ and $B = 1 - x$.
So,$(1 + x)(1 - x) = 1$
$1 - x^2 = 1$
$-x^2 = 0$
$x^2 = 0$
Therefore,$x = 0$.

(B) Given the expression: $S = {\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{zx}}{{yr}}} \right)$.
Using the identity ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B + {\tan ^{ - 1}}C = {\tan ^{ - 1}}\left( {\frac{{A + B + C - ABC}}{{1 - (AB + BC + CA)}}} \right)$,we have:
$A = \frac{{xy}}{{zr}}, B = \frac{{yz}}{{xr}}, C = \frac{{zx}}{{yr}}$.
Then $A + B + C = \frac{{x^2y^2 + y^2z^2 + z^2x^2}}{{xyzr}}$ and $ABC = \frac{{x^2y^2z^2}}{{xyzr^3}} = \frac{{xyz}}{{r^3}}$.
Also,$AB + BC + CA = \frac{{xy^2z}}{{xr^2}} + \frac{{yz^2x}}{{yr^2}} + \frac{{zx^2y}}{{zr^2}} = \frac{{y^2 + z^2 + x^2}}{{r^2}} = \frac{{r^2}}{{r^2}} = 1$.
Since the denominator $1 - (AB + BC + CA) = 1 - 1 = 0$,the argument of ${\tan ^{ - 1}}$ approaches $\infty$.
Therefore,$S = {\tan ^{ - 1}}(\infty) = \frac{\pi }{2}$.

(C) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$.
We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Let $u = \sin^{-1} x$. Then $\cos^{-1} x = \frac{\pi}{2} - u$.
Since $x \in [-1, 1]$,$u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Now,$f(u) = u^3 + (\frac{\pi}{2} - u)^3 = u^3 + \frac{\pi^3}{8} - 3u^2(\frac{\pi}{2}) + 3u(\frac{\pi^2}{4}) - u^3 = \frac{3\pi^2}{4}u - \frac{3\pi}{2}u^2 + \frac{\pi^3}{8}$.
Completing the square: $f(u) = \frac{3\pi}{2} [ -u^2 + \frac{\pi}{2}u ] + \frac{\pi^3}{8} = \frac{3\pi}{2} [ -(u - \frac{\pi}{4})^2 + \frac{\pi^2}{16} ] + \frac{\pi^3}{8} = \frac{3\pi^3}{32} - \frac{3\pi}{2}(u - \frac{\pi}{4})^2 + \frac{\pi^3}{8} = \frac{7\pi^3}{32} - \frac{3\pi}{2}(u - \frac{\pi}{4})^2$.
Wait,let's re-evaluate: $f(u) = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8} = \frac{3\pi}{2}(u - \frac{\pi}{4})^2 + \frac{\pi^3}{32}$.
For $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum occurs at $u = \frac{\pi}{4}$,giving $f_{min} = \frac{\pi^3}{32}$.
The maximum occurs at the boundary $u = -\frac{\pi}{2}$,giving $f_{max} = \frac{3\pi}{2}(-\frac{\pi}{2} - \frac{\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(-\frac{3\pi}{4})^2 + \frac{\pi^3}{32} = \frac{3\pi}{2}(\frac{9\pi^2}{16}) + \frac{\pi^3}{32} = \frac{27\pi^3}{32} + \frac{\pi^3}{32} = \frac{28\pi^3}{32} = \frac{7\pi^3}{8}$.

(A) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$ where $x \in [-1, 1]$.
We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Let $u = \sin^{-1} x$,where $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Then $f(u) = u^3 + (\frac{\pi}{2} - u)^3 = u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8}$.
This is a quadratic in $u$. The vertex is at $u = -\frac{b}{2a} = \frac{3\pi^2/4}{3\pi} = \frac{\pi}{4}$.
Since $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $u = \frac{\pi}{4}$:
$f(\frac{\pi}{4}) = \frac{3\pi}{2}(\frac{\pi^2}{16}) - \frac{3\pi^2}{4}(\frac{\pi}{4}) + \frac{\pi^3}{8} = \frac{3\pi^3}{32} - \frac{3\pi^3}{16} + \frac{\pi^3}{8} = \frac{3\pi^3 - 6\pi^3 + 4\pi^3}{32} = \frac{\pi^3}{32}$.
The maximum value occurs at the boundary $u = -\frac{\pi}{2}$:
$f(-\frac{\pi}{2}) = (-\frac{\pi}{2})^3 + (\pi)^3 = -\frac{\pi^3}{8} + \pi^3 = \frac{7\pi^3}{8}$.
Thus,the range of $f(x)$ is $[\frac{\pi^3}{32}, \frac{7\pi^3}{8}]$.
Given $a < \frac{1}{32}$,we have $a\pi^3 < \frac{\pi^3}{32}$.
Since the minimum value of the expression is $\frac{\pi^3}{32}$,there are no values of $x$ such that $f(x) = a\pi^3$ when $a < \frac{1}{32}$.
Therefore,the number of solutions is $0$.

(A) We know that for $x \in [-1, 1]$,the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ holds true.
Substituting this into the given expression,we get $\frac{\pi}{2} + \tan^{-1}x$.
Since the domain of $\tan^{-1}x$ is $(-\infty, \infty)$,the range of $\tan^{-1}x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Therefore,$-\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2}$.
Adding $\frac{\pi}{2}$ to all parts of the inequality,we get $\frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} + \tan^{-1}x < \frac{\pi}{2} + \frac{\pi}{2}$.
This simplifies to $0 < \frac{\pi}{2} + \tan^{-1}x < \pi$.
Comparing this with $k \le \sin^{-1}x + \cos^{-1}x + \tan^{-1}x \le K$,we find $k = 0$ and $K = \pi$.

(A) We are given the equation: ${({\tan ^{ - 1}}x)^2} + {({\cot ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$.
Using the identity ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$,we can write ${\cot ^{ - 1}}x = \frac{\pi }{2} - {\tan ^{ - 1}}x$.
Substituting this into the equation:
${({\tan ^{ - 1}}x)^2} + {(\frac{\pi }{2} - {\tan ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$.
Let $u = {\tan ^{ - 1}}x$. Then the equation becomes:
$u^2 + (\frac{\pi }{2} - u)^2 = \frac{{5{\pi ^2}}}{8}$.
Expanding the square:
$u^2 + \frac{{{\pi ^2}}}{4} - \pi u + u^2 = \frac{{5{\pi ^2}}}{8}$.
$2u^2 - \pi u + \frac{{{\pi ^2}}}{4} - \frac{{5{\pi ^2}}}{8} = 0$.
$2u^2 - \pi u - \frac{{3{\pi ^2}}}{8} = 0$.
Multiplying by $8$ to clear the denominator:
$16u^2 - 8\pi u - 3{\pi ^2} = 0$.
Solving the quadratic equation for $u$ using the quadratic formula $u = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$:
$u = \frac{{8\pi \pm \sqrt {{{( - 8\pi )}^2} - 4(16)( - 3{\pi ^2})} }}{{32}} = \frac{{8\pi \pm \sqrt {64{\pi ^2} + 192{\pi ^2}} }}{{32}} = \frac{{8\pi \pm \sqrt {256{\pi ^2}} }}{{32}} = \frac{{8\pi \pm 16\pi }}{{32}}$.
So,$u = \frac{{24\pi }}{{32}} = \frac{{3\pi }}{4}$ or $u = \frac{{ - 8\pi }}{{32}} = - \frac{\pi }{4}$.
Since the range of ${\tan ^{ - 1}}x$ is $(-\frac{\pi }{2}, \frac{\pi }{2})$,we must have $u = - \frac{\pi }{4}$.
Therefore,${\tan ^{ - 1}}x = - \frac{\pi }{4} \Rightarrow x = \tan( - \frac{\pi }{4}) = - 1$.

(C) Given that $\tan(x + y) = 33$,we have $x + y = \tan^{-1}(33)$.
Since $x = \tan^{-1}(3)$,we can write $y = \tan^{-1}(33) - x = \tan^{-1}(33) - \tan^{-1}(3)$.
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A - B}{1 + AB}\right)$,we get:
$y = \tan^{-1}\left(\frac{33 - 3}{1 + (33 \times 3)}\right)$.
$y = \tan^{-1}\left(\frac{30}{1 + 99}\right)$.
$y = \tan^{-1}\left(\frac{30}{100}\right)$.
$y = \tan^{-1}(0.3)$.

(C) The given expression is $S = \tan ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}\right) + \tan ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}\right) + \tan ^{ - 1}\left(\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}}\right) + ... + \tan ^{ - 1}\left(\frac{1}{{{c_n}}}\right)$.
First,rewrite the first term: $\tan ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}\right) = \tan ^{ - 1}\left(\frac{\frac{x}{y} - \frac{1}{c_1}}{1 + \frac{x}{y} \cdot \frac{1}{c_1}}\right) = \tan ^{ - 1}\left(\frac{x}{y}\right) - \tan ^{ - 1}\left(\frac{1}{c_1}\right)$.
Now,express the subsequent terms using the identity $\tan ^{ - 1} a - \tan ^{ - 1} b = \tan ^{ - 1}\left(\frac{a - b}{1 + ab}\right)$:
$\tan ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}\right) = \tan ^{ - 1}\left(\frac{1}{c_1}\right) - \tan ^{ - 1}\left(\frac{1}{c_2}\right)$.
$\tan ^{ - 1}\left(\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}}\right) = \tan ^{ - 1}\left(\frac{1}{c_2}\right) - \tan ^{ - 1}\left(\frac{1}{c_3}\right)$.
Continuing this pattern,the general term is $\tan ^{ - 1}\left(\frac{1}{c_{k-1}}\right) - \tan ^{ - 1}\left(\frac{1}{c_k}\right)$.
Summing these up:
$S = \left[\tan ^{ - 1}\left(\frac{x}{y}\right) - \tan ^{ - 1}\left(\frac{1}{c_1}\right)\right] + \left[\tan ^{ - 1}\left(\frac{1}{c_1}\right) - \tan ^{ - 1}\left(\frac{1}{c_2}\right)\right] + ... + \left[\tan ^{ - 1}\left(\frac{1}{c_{n-1}}\right) - \tan ^{ - 1}\left(\frac{1}{c_n}\right)\right] + \tan ^{ - 1}\left(\frac{1}{c_n}\right)$.
This is a telescoping series where all intermediate terms cancel out:
$S = \tan ^{ - 1}\left(\frac{x}{y}\right) - \tan ^{ - 1}\left(\frac{1}{c_n}\right) + \tan ^{ - 1}\left(\frac{1}{c_n}\right) = \tan ^{ - 1}\left(\frac{x}{y}\right)$.

(D) We know that for any $x \in [-1, 1]$,the identity ${{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \frac{\pi }{2}$ holds true.
Given the expression $\sin \left\{ {{\sin }^{ - 1}}\frac{1}{2} + {{\cos }^{ - 1}}\frac{1}{2} \right\}$.
Substituting $x = \frac{1}{2}$ into the identity,we get:
${{\sin }^{ - 1}}\frac{1}{2} + {{\cos }^{ - 1}}\frac{1}{2} = \frac{\pi }{2}$.
Therefore,the expression becomes $\sin \left( \frac{\pi }{2} \right)$.
Since $\sin \left( \frac{\pi }{2} \right) = 1$,the final answer is $1$.

(A) We are given the expression: ${\sin ^{ - 1}}\frac{4}{5} + 2{\tan ^{ - 1}}\frac{1}{3}$.
First,convert ${\sin ^{ - 1}}\frac{4}{5}$ into the $\tan^{-1}$ form. Let ${\sin ^{ - 1}}\frac{4}{5} = \theta$,then $\sin \theta = \frac{4}{5}$.
Using the identity $\tan \theta = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} = \frac{4/5}{\sqrt{1 - 16/25}} = \frac{4/5}{3/5} = \frac{4}{3}$.
So,${\sin ^{ - 1}}\frac{4}{5} = {\tan ^{ - 1}}\frac{4}{3}$.
Next,use the formula $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\frac{2x}{1 - x^2}$ for $2{\tan ^{ - 1}}\frac{1}{3}$:
$2{\tan ^{ - 1}}\frac{1}{3} = {\tan ^{ - 1}}\left( \frac{2(1/3)}{1 - (1/3)^2} \right) = {\tan ^{ - 1}}\left( \frac{2/3}{1 - 1/9} \right) = {\tan ^{ - 1}}\left( \frac{2/3}{8/9} \right) = {\tan ^{ - 1}}\left( \frac{2}{3} \times \frac{9}{8} \right) = {\tan ^{ - 1}}\frac{3}{4}$.
Now,the expression becomes ${\tan ^{ - 1}}\frac{4}{3} + {\tan ^{ - 1}}\frac{3}{4}$.
Since ${\tan ^{ - 1}}x + {\tan ^{ - 1}}\frac{1}{x} = \frac{\pi }{2}$ for $x > 0$,we have ${\tan ^{ - 1}}\frac{4}{3} + {\tan ^{ - 1}}\frac{3}{4} = \frac{\pi }{2}$.
Thus,the correct option is $A$.

(B) The expression $\sin^{-1} x + \cos^{-1} x$ is a fundamental property of inverse trigonometric functions.
For any $x \in [-1, 1]$,the sum of the inverse sine and inverse cosine functions is always equal to $\frac{\pi}{2}$.
Therefore,$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.

(B) We use the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)$ for $xy < 1$.
Given $x = \frac{1}{2}$ and $y = \frac{1}{3}$,we have $xy = \frac{1}{6} < 1$.
Substituting the values:
$\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \tan^{-1} \left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} \right)$
$= \tan^{-1} \left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} \right)$
$= \tan^{-1} \left( \frac{5/6}{5/6} \right)$
$= \tan^{-1} (1)$
$= \frac{\pi}{4}$.

(D) We use the formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ for $xy < 1$.
Given expression: $\tan^{-1}\left(\frac{1}{11}\right) + \tan^{-1}\left(\frac{2}{12}\right)$.
Here $x = \frac{1}{11}$ and $y = \frac{2}{12} = \frac{1}{6}$.
$\tan^{-1}\left(\frac{\frac{1}{11} + \frac{1}{6}}{1 - \frac{1}{11} \times \frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{6+11}{66}}{1 - \frac{1}{66}}\right)$.
$= \tan^{-1}\left(\frac{\frac{17}{66}}{\frac{65}{66}}\right) = \tan^{-1}\left(\frac{17}{65}\right)$.
Since $\tan^{-1}\left(\frac{17}{65}\right)$ is not among the options,the correct choice is $(d)$.

(C) Given equation: ${\tan ^{ - 1}}x + 2{\cot ^{ - 1}}x = \frac{{2\pi }}{3}$
We know that ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}$ for all $x \in \mathbb{R}$.
We can rewrite the given equation as:
$({\tan ^{ - 1}}x + {\cot ^{ - 1}}x) + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$
Substituting the identity:
$\frac{\pi }{2} + {\cot ^{ - 1}}x = \frac{{2\pi }}{3}$
Subtracting $\frac{\pi }{2}$ from both sides:
${\cot ^{ - 1}}x = \frac{{2\pi }}{3} - \frac{\pi }{2}$
${\cot ^{ - 1}}x = \frac{{4\pi - 3\pi }}{6} = \frac{\pi }{6}$
Taking the cotangent of both sides:
$x = \cot \left( \frac{\pi }{6} \right)$
Since $\cot \left( \frac{\pi }{6} \right) = \sqrt 3 $,we get $x = \sqrt 3 $.

(B) Given equation: ${\sin ^{ - 1}}x + {\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{2}$.
We know that ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$.
Also,${\cot ^{ - 1}}\left( {\frac{1}{2}} \right) = {\tan ^{ - 1}}(2)$.
Let ${\tan ^{ - 1}}(2) = \theta$,then $\tan \theta = 2$.
Using the triangle method,if the opposite side is $2$ and the adjacent side is $1$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.
Thus,$\cos \theta = \frac{1}{\sqrt{5}}$,which implies $\theta = {\cos ^{ - 1}}\left( {\frac{1}{\sqrt{5}}} \right)$.
Substituting this into the original equation: ${\sin ^{ - 1}}x + {\cos ^{ - 1}}\left( {\frac{1}{\sqrt{5}}} \right) = \frac{\pi }{2}$.
Comparing this with ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}$,we get $x = \frac{1}{\sqrt{5}}$.

(B) We are given the equation $4\sin^{-1}x + \cos^{-1}x = \pi$.
We know the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.
We can rewrite the given equation as $3\sin^{-1}x + (\sin^{-1}x + \cos^{-1}x) = \pi$.
Substituting the identity,we get $3\sin^{-1}x + \frac{\pi}{2} = \pi$.
Subtracting $\frac{\pi}{2}$ from both sides,we get $3\sin^{-1}x = \frac{\pi}{2}$.
Dividing by $3$,we get $\sin^{-1}x = \frac{\pi}{6}$.
Taking the sine of both sides,we find $x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.

(D) Given $\sin^{-1} C = \sin^{-1} \frac{3}{5} + \cos^{-1} \frac{12}{13}$.
Let $\alpha = \sin^{-1} \frac{3}{5}$ and $\beta = \cos^{-1} \frac{12}{13}$.
Then $\sin \alpha = \frac{3}{5}$,which implies $\cos \alpha = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
And $\cos \beta = \frac{12}{13}$,which implies $\sin \beta = \sqrt{1 - (\frac{12}{13})^2} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.
We have $\sin^{-1} C = \alpha + \beta$,so $C = \sin(\alpha + \beta)$.
Using the identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$:
$C = (\frac{3}{5} \times \frac{12}{13}) + (\frac{4}{5} \times \frac{5}{13})$
$C = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}$.

(B) Let the given expression be $E = \sin \left[ {{\tan }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{2x}}} \right) + {{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) \right]$.
Substitute $x = \tan \theta$,which implies $\theta = \tan^{-1} x$.
The expression becomes:
$E = \sin \left[ {{\tan }^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{2\tan \theta }}} \right) + {{\cos }^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) \right]$
Using trigonometric identities $\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}$ and $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$:
$E = \sin \left[ {{\tan }^{ - 1}}(\cot 2\theta ) + {{\cos }^{ - 1}}(\cos 2\theta ) \right]$
Since $\cot 2\theta = \tan(\frac{\pi}{2} - 2\theta)$:
$E = \sin \left[ {{\tan }^{ - 1}}\left( \tan \left( \frac{\pi}{2} - 2\theta \right) \right) + {{\cos }^{ - 1}}(\cos 2\theta ) \right]$
$E = \sin \left[ \left( \frac{\pi}{2} - 2\theta \right) + 2\theta \right]$
$E = \sin \left( \frac{\pi}{2} \right) = 1$.

(C) Given that ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y + {\cos ^{ - 1}}z = 3\pi $.
We know that the range of the principal value branch of ${\cos ^{ - 1}}x$ is $[0, \pi ]$.
Therefore,$0 \le {\cos ^{ - 1}}x \le \pi $,$0 \le {\cos ^{ - 1}}y \le \pi $,and $0 \le {\cos ^{ - 1}}z \le \pi $.
The sum of three values,each at most $\pi $,can only be $3\pi $ if each individual value is equal to $\pi $.
Thus,${\cos ^{ - 1}}x = \pi $,${\cos ^{ - 1}}y = \pi $,and ${\cos ^{ - 1}}z = \pi $.
This implies $x = \cos \pi = -1$,$y = \cos \pi = -1$,and $z = \cos \pi = -1$.
Substituting these values into the expression $xy + yz + zx$:
$xy + yz + zx = (-1)(-1) + (-1)(-1) + (-1)(-1) = 1 + 1 + 1 = 3$.

(B) We know the identity for inverse trigonometric functions: $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ for $x \in [-1, 1]$.
In the given expression,let $x = -\frac{1}{7}$.
Since $-\frac{1}{7} \in [-1, 1]$,we can apply the identity:
$\cos^{-1}\left( -\frac{1}{7} \right) + \sin^{-1}\left( -\frac{1}{7} \right) = \frac{\pi}{2}$.
Substituting this into the original expression:
$\cos \left[ \cos^{-1}\left( -\frac{1}{7} \right) + \sin^{-1}\left( -\frac{1}{7} \right) \right] = \cos \left( \frac{\pi}{2} \right)$.
Since $\cos \left( \frac{\pi}{2} \right) = 0$,the final result is $0$.

(D) Let $\alpha = \sin^{-1} \left( \frac{3}{5} \right)$. Then $\sin \alpha = \frac{3}{5}$,so $\tan \alpha = \frac{3}{\sqrt{5^2 - 3^2}} = \frac{3}{4}$. Thus,$\alpha = \tan^{-1} \left( \frac{3}{4} \right)$.
Let $\beta = \cos^{-1} \left( \frac{3}{\sqrt{13}} \right)$. Then $\cos \beta = \frac{3}{\sqrt{13}}$,so $\tan \beta = \frac{\sqrt{(\sqrt{13})^2 - 3^2}}{3} = \frac{\sqrt{13-9}}{3} = \frac{2}{3}$. Thus,$\beta = \tan^{-1} \left( \frac{2}{3} \right)$.
Now,the expression becomes $\tan(\alpha + \beta) = \tan \left( \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3} \right)$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan(\alpha + \beta) = \frac{\frac{3}{4} + \frac{2}{3}}{1 - \left( \frac{3}{4} \times \frac{2}{3} \right)} = \frac{\frac{9+8}{12}}{1 - \frac{6}{12}} = \frac{\frac{17}{12}}{\frac{6}{12}} = \frac{17}{6}$.

(D) We use the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$.
Given expression: $\tan \left( \tan^{-1} \frac{1}{2} - \tan^{-1} \frac{1}{3} \right)$.
Applying the formula with $x = \frac{1}{2}$ and $y = \frac{1}{3}$:
$= \tan \left[ \tan^{-1} \left( \frac{\frac{1}{2} - \frac{1}{3}}{1 + (\frac{1}{2} \times \frac{1}{3})} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{\frac{3-2}{6}}{1 + \frac{1}{6}} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{1/6}{7/6} \right) \right]$
$= \tan \left[ \tan^{-1} \left( \frac{1}{7} \right) \right]$
$= \frac{1}{7}$.

(D) Let $\alpha = \cos^{-1} \sqrt{p},$ $\beta = \cos^{-1} \sqrt{1-p},$ and $\gamma = \cos^{-1} \sqrt{1-q}.$
Then $\cos \alpha = \sqrt{p},$ $\cos \beta = \sqrt{1-p},$ and $\cos \gamma = \sqrt{1-q}.$
Consequently,$\sin \alpha = \sqrt{1-p},$ $\sin \beta = \sqrt{p},$ and $\sin \gamma = \sqrt{q}.$
The given equation is $\alpha + \beta + \gamma = \frac{3\pi}{4}.$
This implies $\alpha + \beta = \frac{3\pi}{4} - \gamma.$
Taking cosine on both sides: $\cos(\alpha + \beta) = \cos\left(\frac{3\pi}{4} - \gamma\right).$
$\cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos\left(\pi - \left(\frac{\pi}{4} + \gamma\right)\right) = -\cos\left(\frac{\pi}{4} + \gamma\right).$
$\sqrt{p} \sqrt{1-p} - \sqrt{1-p} \sqrt{p} = -\left(\frac{1}{\sqrt{2}} \cos \gamma - \frac{1}{\sqrt{2}} \sin \gamma\right).$
$0 = -\frac{1}{\sqrt{2}}(\sqrt{1-q} - \sqrt{q}).$
Thus,$\sqrt{1-q} = \sqrt{q}.$
Squaring both sides,$1-q = q,$ which gives $2q = 1,$ or $q = \frac{1}{2}.$

(A) Given: $\cot^{-1}[(\cos \alpha)^{1/2}] - \tan^{-1}[(\cos \alpha)^{1/2}] = x$
Using the identity $\cot^{-1}(y) = \tan^{-1}(\frac{1}{y})$,we get:
$\tan^{-1}[\frac{1}{\sqrt{\cos \alpha}}] - \tan^{-1}[\sqrt{\cos \alpha}] = x$
Using the formula $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}(\frac{A-B}{1+AB})$:
$\tan^{-1}[\frac{\frac{1}{\sqrt{\cos \alpha}} - \sqrt{\cos \alpha}}{1 + (\frac{1}{\sqrt{\cos \alpha}})(\sqrt{\cos \alpha})}] = x$
$\tan^{-1}[\frac{\frac{1-\cos \alpha}{\sqrt{\cos \alpha}}}{1+1}] = x$
$\tan x = \frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}$
Now,we need to find $\sin x$. Using the identity $\sin x = \frac{\tan x}{\sqrt{1+\tan^2 x}}$ or by constructing a right triangle where the opposite side is $1-\cos \alpha$ and the adjacent side is $2\sqrt{\cos \alpha}$:
Hypotenuse = $\sqrt{(1-\cos \alpha)^2 + (2\sqrt{\cos \alpha})^2} = \sqrt{1 - 2\cos \alpha + \cos^2 \alpha + 4\cos \alpha} = \sqrt{1 + 2\cos \alpha + \cos^2 \alpha} = \sqrt{(1+\cos \alpha)^2} = 1+\cos \alpha$
Therefore,$\sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = \tan^2(\frac{\alpha}{2})$.

(B) Given: ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi $
Taking tangent on both sides:
${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi - {\tan ^{ - 1}}z$
Applying the formula ${\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( \frac{A+B}{1-AB} \right)$:
${\tan ^{ - 1}}\left( \frac{x+y}{1-xy} \right) = \pi - {\tan ^{ - 1}}z$
Taking tangent on both sides:
$\frac{x+y}{1-xy} = \tan(\pi - {\tan ^{ - 1}}z)$
Since $\tan(\pi - \theta) = -\tan \theta$:
$\frac{x+y}{1-xy} = -z$
$x + y = -z(1 - xy)$
$x + y = -z + xyz$
$x + y + z = xyz$
Dividing both sides by $xyz$:
$\frac{x}{xyz} + \frac{y}{xyz} + \frac{z}{xyz} = \frac{xyz}{xyz}$
$\frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} = 1$
Thus,$\frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}} = 1$.

(C) Let $a = \tan \theta$. Then $\theta = \tan^{-1}(a)$.
The expression becomes:
$\tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)} \right]$
Using the trigonometric identities $\sin 2\theta = \frac{2\tan \theta}{1 + \tan^2 \theta}$ and $\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$:
$= \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}(\sin 2\theta ) + \frac{1}{2}{{\cos }^{ - 1}}(\cos 2\theta )} \right]$
Assuming the principal value range for $2\theta$:
$= \tan \left[ {\frac{1}{2}(2\theta ) + \frac{1}{2}(2\theta )} \right]$
$= \tan (\theta + \theta) = \tan 2\theta$
Using the formula $\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}$:
$= \frac{2a}{1 - a^2}$

(C) Given equation: $\cos (2\sin ^{ - 1}x) = \frac{1}{9}$
Let $\sin ^{ - 1}x = \theta,$ then $\sin \theta = x.$
The equation becomes $\cos (2\theta) = \frac{1}{9}.$
Using the identity $\cos (2\theta) = 1 - 2\sin ^2\theta,$ we have:
$1 - 2\sin ^2\theta = \frac{1}{9}$
$1 - 2x^2 = \frac{1}{9}$
$2x^2 = 1 - \frac{1}{9} = \frac{8}{9}$
$x^2 = \frac{4}{9}$
$x = \pm \frac{2}{3}.$
Since the range of $\sin ^{ - 1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}],$ the value of $2\sin ^{ - 1}x$ lies in $[-\pi, \pi].$ The cosine function is defined for both positive and negative values in this range,so both $x = \frac{2}{3}$ and $x = -\frac{2}{3}$ are valid solutions.

(B) Given the equation: $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$.
Using the formula $2\tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,we get:
$\tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}(2\csc x)$.
Since $1-\cos^2 x = \sin^2 x$,this simplifies to:
$\frac{2\cos x}{\sin^2 x} = 2\csc x$.
Substituting $\csc x = \frac{1}{\sin x}$:
$\frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x}$.
Assuming $\sin x \neq 0$,we multiply both sides by $\sin^2 x$:
$2\cos x = 2\sin x$.
Dividing by $2\cos x$ (assuming $\cos x \neq 0$):
$\tan x = 1$.
Thus,$x = \frac{\pi}{4}$.

(D) We use the formula $2\tan^{-1}(x) = \tan^{-1}\left( \frac{2x}{1-x^2} \right)$.
First,calculate $2\tan^{-1}\left( \frac{1}{5} \right)$:
$2\tan^{-1}\left( \frac{1}{5} \right) = \tan^{-1}\left( \frac{2(1/5)}{1-(1/5)^2} \right) = \tan^{-1}\left( \frac{2/5}{1-1/25} \right) = \tan^{-1}\left( \frac{2/5}{24/25} \right) = \tan^{-1}\left( \frac{2}{5} \times \frac{25}{24} \right) = \tan^{-1}\left( \frac{5}{12} \right)$.
Now,substitute this into the expression:
$\tan \left[ \tan^{-1}\left( \frac{5}{12} \right) - \frac{\pi}{4} \right] = \tan \left[ \tan^{-1}\left( \frac{5}{12} \right) - \tan^{-1}(1) \right]$.
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left( \frac{x-y}{1+xy} \right)$:
$\tan \left[ \tan^{-1}\left( \frac{5/12 - 1}{1 + (5/12)(1)} \right) \right] = \frac{5/12 - 1}{1 + 5/12} = \frac{-7/12}{17/12} = -\frac{7}{17}$.

(B) Let $x = {\tan ^2}\theta$,which implies $\sqrt{x} = \tan \theta$,so $\theta = {\tan ^{ - 1}}\sqrt x $.
Substituting $x = {\tan ^2}\theta$ into the expression:
$\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$
Using the trigonometric identity $\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$:
$= \frac{1}{2}{\cos ^{ - 1}}(\cos 2\theta)$
$= \frac{1}{2}(2\theta) = \theta$
Substituting back $\theta = {\tan ^{ - 1}}\sqrt x $:
$= {\tan ^{ - 1}}\sqrt x $.
Thus,the correct option is $B$.

(B) Let $\theta = \tan^{-1} \frac{1}{3}$,then $\tan \theta = \frac{1}{3}$.
We need to find $\sin(4\theta)$.
First,calculate $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} = \frac{2(1/3)}{1 - (1/9)} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4}$.
Now,calculate $\sin(4\theta) = \sin(2(2\theta)) = \frac{2 \tan(2\theta)}{1 + \tan^2(2\theta)}$.
Substituting $\tan(2\theta) = \frac{3}{4}$:
$\sin(4\theta) = \frac{2(3/4)}{1 + (3/4)^2} = \frac{3/2}{1 + 9/16} = \frac{3/2}{25/16} = \frac{3}{2} \times \frac{16}{25} = \frac{24}{25}$.
Thus,the correct option is $B$.

(D) We know the trigonometric identity for $\tan 3\theta$ is given by:
$\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$
Let $\tan \theta = a$,then $\theta = \tan^{-1} a$.
Substituting this into the identity,we get:
$\tan 3(\tan^{-1} a) = \frac{3a - a^3}{1 - 3a^2}$
Taking $\tan^{-1}$ on both sides,we obtain:
$3 \tan^{-1} a = \tan^{-1} \left( \frac{3a - a^3}{1 - 3a^2} \right)$
Thus,the correct option is $D$.

(B) Given the equation: $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$.
Substitute $x = \tan \theta$,which implies $\theta = {\tan ^{ - 1}}x$.
Using standard trigonometric identities:
$\sin 2\theta = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$,$\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$,and $\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$.
The equation becomes:
$3(2\theta ) - 4(2\theta ) + 2(2\theta ) = \frac{\pi }{3}$.
Simplifying the expression:
$6\theta - 8\theta + 4\theta = \frac{\pi }{3}$.
$2\theta = \frac{\pi }{3} \Rightarrow \theta = \frac{\pi }{6}$.
Since $\theta = {\tan ^{ - 1}}x$,we have ${\tan ^{ - 1}}x = \frac{\pi }{6}$.
Therefore,$x = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$.

(B) Let the expression be $E = \sin \left( {2{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) + \cos ({\tan ^{ - 1}}(2\sqrt 2 ))$.
First,use the formula $2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$.
For $x = \frac{1}{3}$,we have $2\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2/3}{1-1/9}\right) = \tan^{-1}\left(\frac{2/3}{8/9}\right) = \tan^{-1}\left(\frac{3}{4}\right)$.
Now,$\sin(\tan^{-1}(3/4))$: If $\tan(\theta) = 3/4$,then $\sin(\theta) = 3/5$.
Next,for $\cos(\tan^{-1}(2\sqrt{2}))$: If $\tan(\phi) = 2\sqrt{2}$,then $\sec^2(\phi) = 1 + \tan^2(\phi) = 1 + (2\sqrt{2})^2 = 1 + 8 = 9$.
So,$\sec(\phi) = 3$,which means $\cos(\phi) = 1/3$.
Adding these values: $E = \frac{3}{5} + \frac{1}{3} = \frac{9+5}{15} = \frac{14}{15}$.

(D) Let $f(x) = \cos ^{ - 1}\left( \frac{3 + 5\cos x}{5 + 3\cos x} \right)$.
Using the formula $\cos x = \frac{1 - \tan ^2(x/2)}{1 + \tan ^2(x/2)}$,we substitute:
$\frac{3 + 5\left( \frac{1 - \tan ^2(x/2)}{1 + \tan ^2(x/2)} \right)}{5 + 3\left( \frac{1 - \tan ^2(x/2)}{1 + \tan ^2(x/2)} \right)} = \frac{3(1 + \tan ^2(x/2)) + 5(1 - \tan ^2(x/2))}{5(1 + \tan ^2(x/2)) + 3(1 - \tan ^2(x/2))}$
$= \frac{3 + 3\tan ^2(x/2) + 5 - 5\tan ^2(x/2)}{5 + 5\tan ^2(x/2) + 3 - 3\tan ^2(x/2)} = \frac{8 - 2\tan ^2(x/2)}{8 + 2\tan ^2(x/2)} = \frac{4 - \tan ^2(x/2)}{4 + \tan ^2(x/2)}$.
Let $t = \tan(x/2)$. Then the expression is $\cos ^{ - 1}\left( \frac{4 - t^2}{4 + t^2} \right) = \cos ^{ - 1}\left( \frac{1 - (t/2)^2}{1 + (t/2)^2} \right)$.
Using the identity $\cos ^{ - 1}\left( \frac{1 - u^2}{1 + u^2} \right) = 2\tan ^{ - 1}u$ (for $u \ge 0$),we get $2\tan ^{ - 1}\left( \frac{t}{2} \right) = 2\tan ^{ - 1}\left( \frac{1}{2}\tan \frac{x}{2} \right)$.

(A) Given the equation: $\cos^{-1} x - \cos^{-1} \frac{y}{2} = \alpha$.
Let $\cos^{-1} x = A$ and $\cos^{-1} \frac{y}{2} = B$.
Then $x = \cos A$ and $\frac{y}{2} = \cos B$,so $y = 2 \cos B$.
The given equation becomes $A - B = \alpha$.
Taking cosine on both sides: $\cos(A - B) = \cos \alpha$.
Using the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we get:
$x \cdot \frac{y}{2} + \sqrt{1 - x^2} \sqrt{1 - (\frac{y}{2})^2} = \cos \alpha$.
$\sqrt{1 - x^2} \sqrt{1 - \frac{y^2}{4}} = \cos \alpha - \frac{xy}{2}$.
Squaring both sides: $(1 - x^2)(1 - \frac{y^2}{4}) = (\cos \alpha - \frac{xy}{2})^2$.
$1 - \frac{y^2}{4} - x^2 + \frac{x^2 y^2}{4} = \cos^2 \alpha - xy \cos \alpha + \frac{x^2 y^2}{4}$.
$1 - x^2 - \frac{y^2}{4} = \cos^2 \alpha - xy \cos \alpha$.
Multiply the entire equation by $4$:
$4 - 4x^2 - y^2 = 4 \cos^2 \alpha - 4xy \cos \alpha$.
Rearranging the terms:
$4x^2 - 4xy \cos \alpha + y^2 = 4 - 4 \cos^2 \alpha$.
$4x^2 - 4xy \cos \alpha + y^2 = 4(1 - \cos^2 \alpha)$.
Since $1 - \cos^2 \alpha = \sin^2 \alpha$,we get:
$4x^2 - 4xy \cos \alpha + y^2 = 4 \sin^2 \alpha$.

(B) Given the equation: $\tan^{-1} x + \tan^{-1} y = \frac{\pi}{4}$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x + y}{1 - xy} \right)$,we get:
$\tan^{-1} \left( \frac{x + y}{1 - xy} \right) = \frac{\pi}{4}$.
Taking the tangent of both sides:
$\frac{x + y}{1 - xy} = \tan \left( \frac{\pi}{4} \right)$.
Since $\tan \left( \frac{\pi}{4} \right) = 1$,we have:
$\frac{x + y}{1 - xy} = 1$.
Multiplying both sides by $(1 - xy)$,we get:
$x + y = 1 - xy$.
Rearranging the terms,we obtain:
$x + y + xy = 1$.

(C) Given the equation: $\sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$
Rearranging the terms,we get: $\sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x$
Taking $\sin$ on both sides: $1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}x\right)$
Using the identity $\sin(\frac{\pi}{2} + \theta) = \cos \theta$: $1 - x = \cos(2\sin^{-1}x)$
Using the identity $\cos(2\theta) = 1 - 2\sin^2\theta$: $1 - x = 1 - 2\sin^2(\sin^{-1}x)$
Simplifying: $1 - x = 1 - 2x^2$
$2x^2 - x = 0$
$x(2x - 1) = 0$
So,$x = 0$ or $x = 1/2$.
Checking $x = 1/2$: $\sin^{-1}(1 - 1/2) - 2\sin^{-1}(1/2) = \sin^{-1}(1/2) - 2(\pi/6) = \pi/6 - \pi/3 = -\pi/6 \neq \pi/2$.
Checking $x = 0$: $\sin^{-1}(1 - 0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \pi/2$.
Thus,the only solution is $x = 0$.

(C) Given that $\angle A = 90^\circ$ in $\triangle ABC$,by the Pythagorean theorem,we have $a^2 = b^2 + c^2$,where $a$ is the hypotenuse.
We need to evaluate $\tan^{-1}\left(\frac{c}{a+b}\right) + \tan^{-1}\left(\frac{b}{a+c}\right)$.
Using the formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$,we get:
$= \tan^{-1}\left[ \frac{\frac{c}{a+b} + \frac{b}{a+c}}{1 - \left(\frac{c}{a+b}\right)\left(\frac{b}{a+c}\right)} \right]$
$= \tan^{-1}\left[ \frac{c(a+c) + b(a+b)}{(a+b)(a+c) - bc} \right]$
$= \tan^{-1}\left[ \frac{ac + c^2 + ab + b^2}{a^2 + ac + ab + bc - bc} \right]$
Since $b^2 + c^2 = a^2$,the numerator becomes $ac + ab + a^2$.
$= \tan^{-1}\left[ \frac{a^2 + ab + ac}{a^2 + ab + ac} \right]$
$= \tan^{-1}(1) = \frac{\pi}{4}$.

(D) Given equation: $\sin^{-1} x - \sin^{-1} 2x = \pm \frac{\pi}{3}$.
Taking $\sin$ on both sides:
$\sin(\sin^{-1} x - \sin^{-1} 2x) = \sin(\pm \frac{\pi}{3}) = \pm \frac{\sqrt{3}}{2}$.
Using the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$,where $A = \sin^{-1} x$ and $B = \sin^{-1} 2x$:
$x \sqrt{1 - (2x)^2} - 2x \sqrt{1 - x^2} = \pm \frac{\sqrt{3}}{2}$.
$x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2} = \pm \frac{\sqrt{3}}{2}$.
Squaring both sides:
$(x \sqrt{1 - 4x^2} - 2x \sqrt{1 - x^2})^2 = \frac{3}{4}$.
$x^2(1 - 4x^2) + 4x^2(1 - x^2) - 4x^2 \sqrt{(1 - 4x^2)(1 - x^2)} = \frac{3}{4}$.
$x^2 - 4x^4 + 4x^2 - 4x^4 - 4x^2 \sqrt{1 - 5x^2 + 4x^4} = \frac{3}{4}$.
$5x^2 - 8x^4 - \frac{3}{4} = 4x^2 \sqrt{1 - 5x^2 + 4x^4}$.
Substituting $x^2 = \frac{1}{4}$:
$5(\frac{1}{4}) - 8(\frac{1}{16}) - \frac{3}{4} = \frac{5}{4} - \frac{1}{2} - \frac{3}{4} = 0$.
$4(\frac{1}{4}) \sqrt{1 - 5(\frac{1}{4}) + 4(\frac{1}{16})} = 1 \sqrt{1 - \frac{5}{4} + \frac{1}{4}} = 1 \sqrt{0} = 0$.
Since $0 = 0$,$x = \pm \frac{1}{2}$ is the correct solution.

(A) We use the identity $3 \sin^{-1}(x) = \sin^{-1}(3x - 4x^3)$ for $|x| \leq \frac{1}{2}$.
Here,$x = \frac{1}{5}$,which satisfies the condition.
$\sin \left[ 3 \sin^{-1} \left( \frac{1}{5} \right) \right] = \sin \left[ \sin^{-1} \left( 3 \left( \frac{1}{5} \right) - 4 \left( \frac{1}{5} \right)^3 \right) \right]$
$= \sin \left[ \sin^{-1} \left( \frac{3}{5} - \frac{4}{125} \right) \right]$
$= \sin \left[ \sin^{-1} \left( \frac{75 - 4}{125} \right) \right]$
$= \sin \left[ \sin^{-1} \left( \frac{71}{125} \right) \right] = \frac{71}{125}$.

(D) Let $y = \cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right]$.
Rationalizing the denominator:
$y = \cot ^{ - 1}\left[ \frac{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x})^2}{(\sqrt {1 - \sin x})^2 - (\sqrt {1 + \sin x})^2} \right]$
Expanding the numerator and denominator:
$y = \cot ^{ - 1}\left[ \frac{(1 - \sin x) + (1 + \sin x) + 2\sqrt {(1 - \sin x)(1 + \sin x)}}{(1 - \sin x) - (1 + \sin x)} \right]$
$y = \cot ^{ - 1}\left[ \frac{2 + 2\sqrt {1 - \sin ^2 x}}{-2\sin x} \right] = \cot ^{ - 1}\left[ \frac{2(1 + \cos x)}{-2\sin x} \right]$
Using half-angle formulas $1 + \cos x = 2\cos ^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$:
$y = \cot ^{ - 1}\left[ \frac{2(2\cos ^2(x/2))}{-2(2\sin(x/2)\cos(x/2))} \right] = \cot ^{ - 1}\left[ -\frac{\cos(x/2)}{\sin(x/2)} \right]$
$y = \cot ^{ - 1}\left( -\cot(x/2) \right) = \cot ^{ - 1}\left( \cot(\pi - x/2) \right) = \pi - \frac{x}{2}$.

(C) Given that $\theta = \tan^{-1} a$ and $\phi = \tan^{-1} b$ with $ab = -1$.
From the given equations,we have $\tan \theta = a$ and $\tan \phi = b$.
Substituting these into the condition $ab = -1$,we get $\tan \theta \tan \phi = -1$.
This implies $\tan \theta = -\frac{1}{\tan \phi} = -\cot \phi$.
We know that $-\cot \phi = \tan(\phi + \frac{\pi}{2})$ or $\tan(\phi - \frac{\pi}{2})$.
Since the range of the principal value branch of $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,the difference $\theta - \phi$ results in $\pm \frac{\pi}{2}$.
Specifically,$\theta - \phi = \tan^{-1} a - \tan^{-1} b$. Using the identity $\tan^{-1} a - \tan^{-1} b = \tan^{-1}(\frac{a-b}{1+ab})$,since $ab = -1$,the denominator becomes $0$,which corresponds to the value $\frac{\pi}{2}$ or $-\frac{\pi}{2}$.
Thus,$\theta - \phi = \pm \frac{\pi}{2}$.

(A) We know the identity for inverse trigonometric functions: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.
Given that $\sin^{-1} x = \frac{\pi}{5}$.
Substituting this value into the identity,we get: $\frac{\pi}{5} + \cos^{-1} x = \frac{\pi}{2}$.
Therefore,$\cos^{-1} x = \frac{\pi}{2} - \frac{\pi}{5}$.
Taking the least common multiple $(LCM)$ of $2$ and $5$,which is $10$,we get: $\cos^{-1} x = \frac{5\pi - 2\pi}{10} = \frac{3\pi}{10}$.

(B) Given that $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$.
Let $\cos^{-1} p = A$,$\cos^{-1} q = B$,and $\cos^{-1} r = C$.
Then $A + B + C = \pi$,which implies $A + B = \pi - C$.
Taking $\cos$ on both sides: $\cos(A + B) = \cos(\pi - C) = -\cos C$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$,we get:
$pq - \sqrt{1 - p^2} \sqrt{1 - q^2} = -r$.
$pq + r = \sqrt{1 - p^2} \sqrt{1 - q^2}$.
Squaring both sides: $(pq + r)^2 = (1 - p^2)(1 - q^2)$.
$p^2q^2 + r^2 + 2pqr = 1 - p^2 - q^2 + p^2q^2$.
$p^2 + q^2 + r^2 + 2pqr = 1$.

(B) Let $\theta = \frac{1}{2}{\cos ^{ - 1}}\frac{a}{b}$.
Then,$2\theta = {\cos ^{ - 1}}\frac{a}{b}$,which implies $\cos 2\theta = \frac{a}{b}$.
The expression becomes $\tan \left( {\frac{\pi }{4} + \theta } \right) + \tan \left( {\frac{\pi }{4} - \theta } \right)$.
Using the formula $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan (A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ with $\tan \frac{\pi }{4} = 1$:
$= \frac{1 + \tan \theta }{1 - \tan \theta } + \frac{1 - \tan \theta }{1 + \tan \theta }$
$= \frac{{(1 + \tan \theta )^2 + (1 - \tan \theta )^2}}{{(1 - \tan \theta )(1 + \tan \theta )}}$
$= \frac{{1 + {{\tan }^2}\theta + 2\tan \theta + 1 + {{\tan }^2}\theta - 2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$= \frac{{2(1 + {{\tan }^2}\theta )}}{{1 - {{\tan }^2}\theta }}$
$= \frac{2}{{\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}}}$
$= \frac{2}{{\cos 2\theta }}$
Since $\cos 2\theta = \frac{a}{b}$,the expression equals $\frac{2}{{a/b}} = \frac{{2b}}{a}$.