Properties of ITF Questions in English

(B) We are given the equation: $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$.
Using the identity $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$ and $\cos ^{-1} y = \frac{\pi}{2} - \sin ^{-1} y$,we substitute these into the equation:
$(\frac{\pi}{2} - \cos ^{-1} x) + (\frac{\pi}{2} - \sin ^{-1} y) = \frac{3 \pi}{10}$.
Simplifying the left side:
$\pi - (\cos ^{-1} x + \sin ^{-1} y) = \frac{3 \pi}{10}$.
Rearranging the terms to solve for $\cos ^{-1} x + \sin ^{-1} y$:
$\cos ^{-1} x + \sin ^{-1} y = \pi - \frac{3 \pi}{10}$.
$\cos ^{-1} x + \sin ^{-1} y = \frac{10\pi - 3\pi}{10} = \frac{7 \pi}{10}$.

(C) We know that $\sin \left(\cot ^{-1} x\right) = \sin \left(\sin ^{-1} \left(\frac{1}{\sqrt{1+x^2}}\right)\right) = \frac{1}{\sqrt{1+x^2}}$.
Also,$\cos \left(\tan ^{-1}(1+x)\right) = \cos \left(\cos ^{-1} \left(\frac{1}{\sqrt{1+(1+x)^2}}\right)\right) = \frac{1}{\sqrt{1+(1+x)^2}}$.
Given the equation $\sin \left(\cot ^{-1}(x)\right) = \cos \left(\tan ^{-1}(1+x)\right)$,we equate the two expressions:
$\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+(1+x)^2}}$.
Squaring both sides,we get $1+x^2 = 1+(1+x)^2$.
$1+x^2 = 1 + 1 + 2x + x^2$.
$1+x^2 = 2 + 2x + x^2$.
Subtracting $x^2$ from both sides,we get $1 = 2 + 2x$.
$-1 = 2x$.
$x = -\frac{1}{2}$.

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

(B) We know that $\sin ^{-1}\left(\frac{3}{5}\right) = \tan ^{-1}\left(\frac{3}{\sqrt{1^2 - (3/5)^2}}\right) = \tan ^{-1}\left(\frac{3/5}{4/5}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$,we get:
$\tan \left[\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right]$
$= \tan \left[\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)\right]$
$= \tan \left[\tan ^{-1}\left(\frac{\frac{9+8}{12}}{1-\frac{6}{12}}\right)\right]$
$= \tan \left[\tan ^{-1}\left(\frac{17/12}{6/12}\right)\right]$
$= \tan \left[\tan ^{-1}\left(\frac{17}{6}\right)\right]$
$= \frac{17}{6}$

(C) Given the equation $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$.
Using the identity $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^2}\sqrt{1-v^2})$,we have $\cos ^{-1} (\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}}) = \alpha$.
Taking cosine on both sides: $\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha$.
Rearranging: $\sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha - \frac{xy}{3}$.
Squaring both sides: $(1-x^2)(1-\frac{y^2}{9}) = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$.
$1 - \frac{y^2}{9} - x^2 + \frac{x^2 y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$.
$1 - x^2 - \frac{y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha$.
$1 - \cos^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$.
$\sin^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$.
Multiplying by $9$: $9 \sin^2 \alpha = 9x^2 - 6xy \cos \alpha + y^2$.

(D) Let $\theta = \cot^{-1} x$,then $\cot \theta = x$. Since $\cot \theta = \frac{x}{1}$,we have $\sin \theta = \frac{1}{\sqrt{1+x^2}}$.
Thus,$\sin(\cot^{-1} x) = \frac{1}{\sqrt{1+x^2}}$.
Now,let $\phi = \tan^{-1} \left( \frac{1}{\sqrt{1+x^2}} \right)$,then $\tan \phi = \frac{1}{\sqrt{1+x^2}}$.
We need to find $\cos \phi$. Using the identity $\sec^2 \phi = 1 + \tan^2 \phi$,we get $\sec^2 \phi = 1 + \frac{1}{1+x^2} = \frac{1+x^2+1}{1+x^2} = \frac{x^2+2}{x^2+1}$.
Therefore,$\cos^2 \phi = \frac{x^2+1}{x^2+2}$,which implies $\cos \phi = \sqrt{\frac{x^2+1}{x^2+2}}$.

(A) Given that in $\Delta ABC$,$C=90^{\circ}$,therefore by Pythagoras theorem,$a^{2}+b^{2}=c^{2}$.
We need to evaluate $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$:
$= \tan ^{-1} \left[ \frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \left( \frac{a}{b+c} \right) \left( \frac{b}{c+a} \right)} \right]$
$= \tan ^{-1} \left[ \frac{a(c+a) + b(b+c)}{(b+c)(c+a) - ab} \right]$
$= \tan ^{-1} \left[ \frac{ac + a^{2} + b^{2} + bc}{bc + ab + c^{2} + ac - ab} \right]$
Since $a^{2}+b^{2}=c^{2}$,the numerator becomes $ac + c^{2} + bc = c(a+b+c)$.
The denominator becomes $bc + ac + c^{2} = c(b+a+c)$.
$= \tan ^{-1} \left( \frac{c(a+b+c)}{c(a+b+c)} \right)$
$= \tan ^{-1}(1) = \frac{\pi}{4}$.

(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$
We can rewrite this as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)\right)$
Using the property $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$,the expression becomes:
$\cos \left(\cos ^{-1} x+\frac{\pi}{2}\right)$
Using the trigonometric identity $\cos \left(\theta+\frac{\pi}{2}\right) = -\sin \theta$:
$= -\sin \left(\cos ^{-1} x\right)$
Since $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$,we have:
$= -\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) = -\sqrt{1-x^2}$
Substituting $x=\frac{1}{5}$:
$= -\sqrt{1-\left(\frac{1}{5}\right)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{2 \sqrt{6}}{5}$

(C) Given: $\tan ^{-1} x + \tan ^{-1} y + \tan ^{-1} z = \frac{\pi}{2}$
$\tan ^{-1} x + \tan ^{-1} y = \frac{\pi}{2} - \tan ^{-1} z$
Using the identity $\frac{\pi}{2} - \tan ^{-1} z = \cot ^{-1} z = \tan ^{-1} (\frac{1}{z})$:
$\tan ^{-1} (\frac{x+y}{1-xy}) = \tan ^{-1} (\frac{1}{z})$
Comparing both sides:
$\frac{x+y}{1-xy} = \frac{1}{z}$
$z(x+y) = 1 - xy$
$zx + zy = 1 - xy$
$xy + yz + zx = 1$

(A) Given the equation: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
Taking $\sin ^{-1}$ on both sides:
$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \sin ^{-1}(1)$
Since $\sin ^{-1}(1) = \frac{\pi}{2}$,we have:
$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$
We know the identity $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in [-1, 1]$.
Comparing the two equations,we get:
$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$
Therefore,$x = \frac{1}{5}$.

(D) Given that $\tan ^{-1}\left(\frac{x}{2}\right)+\tan ^{-1}\left(\frac{y}{2}\right)+\tan ^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}$.
Let $A = \tan ^{-1}\left(\frac{x}{2}\right)$,$B = \tan ^{-1}\left(\frac{y}{2}\right)$,and $C = \tan ^{-1}\left(\frac{z}{2}\right)$.
Then $A+B+C = \frac{\pi}{2}$,which implies $A+B = \frac{\pi}{2}-C$.
Taking $\tan$ on both sides,we get $\tan(A+B) = \tan\left(\frac{\pi}{2}-C\right) = \cot(C) = \frac{1}{\tan(C)}$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\frac{\frac{x}{2} + \frac{y}{2}}{1 - \frac{x}{2} \cdot \frac{y}{2}} = \frac{1}{z/2}$.
This simplifies to $\frac{(x+y)/2}{(4-xy)/4} = \frac{2}{z}$,which is $\frac{2(x+y)}{4-xy} = \frac{2}{z}$.
Dividing by $2$,we get $\frac{x+y}{4-xy} = \frac{1}{z}$.
Cross-multiplying gives $z(x+y) = 4-xy$,which means $zx + zy = 4 - xy$.
Rearranging the terms,we get $xy + yz + zx = 4$.

(B) Given the equation: $\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}+\sin ^{-1} x=\frac{\pi}{2}$.
We know that $\sin ^{-1} x = \cos ^{-1} \sqrt{1-x^2}$.
So,$\sin ^{-1} \frac{1}{3} = \cos ^{-1} \sqrt{1-(\frac{1}{3})^2} = \cos ^{-1} \sqrt{\frac{8}{9}} = \cos ^{-1} \frac{2\sqrt{2}}{3}$.
And $\sin ^{-1} \frac{3}{5} = \cos ^{-1} \sqrt{1-(\frac{3}{5})^2} = \cos ^{-1} \sqrt{\frac{16}{25}} = \cos ^{-1} \frac{4}{5}$.
Substituting these into the equation: $\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5} + \sin ^{-1} x = \frac{\pi}{2}$.
$\sin ^{-1} x = \frac{\pi}{2} - (\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5})$.
Using $\frac{\pi}{2} - \cos ^{-1} \theta = \sin ^{-1} \theta$,we get $\sin ^{-1} x = \sin ^{-1} (\cos ^{-1} \frac{2\sqrt{2}}{3} + \cos ^{-1} \frac{4}{5})$.
Let $\alpha = \cos ^{-1} \frac{2\sqrt{2}}{3}$ and $\beta = \cos ^{-1} \frac{4}{5}$. Then $\cos \alpha = \frac{2\sqrt{2}}{3}$ and $\cos \beta = \frac{4}{5}$.
Then $\sin \alpha = \sqrt{1-(\frac{2\sqrt{2}}{3})^2} = \frac{1}{3}$ and $\sin \beta = \sqrt{1-(\frac{4}{5})^2} = \frac{3}{5}$.
We need $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = (\frac{1}{3} \times \frac{4}{5}) + (\frac{2\sqrt{2}}{3} \times \frac{3}{5}) = \frac{4}{15} + \frac{6\sqrt{2}}{15} = \frac{4+6\sqrt{2}}{15}$.
Wait,re-evaluating: $\sin ^{-1} x = \frac{\pi}{2} - (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5}) = \cos ^{-1} (\sin ^{-1} \frac{1}{3} + \sin ^{-1} \frac{3}{5})$.
Let $A = \sin ^{-1} \frac{1}{3}$ and $B = \sin ^{-1} \frac{3}{5}$. Then $\sin A = \frac{1}{3}, \cos A = \frac{2\sqrt{2}}{3}$ and $\sin B = \frac{3}{5}, \cos B = \frac{4}{5}$.
$\sin(A+B) = \sin A \cos B + \cos A \sin B = (\frac{1}{3} \times \frac{4}{5}) + (\frac{2\sqrt{2}}{3} \times \frac{3}{5}) = \frac{4+6\sqrt{2}}{15}$.
Since $\sin ^{-1} x = \frac{\pi}{2} - (A+B)$,then $x = \sin(\frac{\pi}{2} - (A+B)) = \cos(A+B) = \cos A \cos B - \sin A \sin B$.
$x = (\frac{2\sqrt{2}}{3} \times \frac{4}{5}) - (\frac{1}{3} \times \frac{3}{5}) = \frac{8\sqrt{2}-3}{15}$.
Thus,the correct option is $B$.

(B) Let $\theta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)$.
Then $2\theta = \cos ^{-1}\left(\frac{a}{b}\right)$,which implies $\cos 2\theta = \frac{a}{b}$.
The given expression is $\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)$.
Using the formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,where $\tan(\frac{\pi}{4}) = 1$:
$= \frac{1+\tan \theta}{1-\tan \theta} + \frac{1-\tan \theta}{1+\tan \theta}$
$= \frac{(1+\tan \theta)^2 + (1-\tan \theta)^2}{(1-\tan \theta)(1+\tan \theta)}$
$= \frac{1 + 2\tan \theta + \tan^2 \theta + 1 - 2\tan \theta + \tan^2 \theta}{1 - \tan^2 \theta}$
$= \frac{2(1+\tan^2 \theta)}{1-\tan^2 \theta}$
$= \frac{2}{\frac{1-\tan^2 \theta}{1+\tan^2 \theta}}$
$= \frac{2}{\cos 2\theta}$
Substituting $\cos 2\theta = \frac{a}{b}$,we get:
$= \frac{2}{a/b} = \frac{2b}{a}$.

(D) We know that the range of $\cos ^{-1} \theta$ is $[0, \pi]$.
Given that $\cos ^{-1} x + \cos ^{-1} y + \cos ^{-1} z = 3\pi$.
Since each term can be at most $\pi$,the only way their sum can be $3\pi$ is if each term is exactly $\pi$.
Therefore,$\cos ^{-1} x = \pi$,$\cos ^{-1} y = \pi$,and $\cos ^{-1} z = \pi$.
This implies $x = \cos(\pi) = -1$,$y = \cos(\pi) = -1$,and $z = \cos(\pi) = -1$.
Now,substitute these values into the expression $x^2 + y^2 + z^2 - 2xyz$:
$(-1)^2 + (-1)^2 + (-1)^2 - 2(-1)(-1)(-1)$
$= 1 + 1 + 1 - 2(-1)$
$= 3 + 2$
$= 5$.

(C) Given: $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$
Using the property $\cot ^{-1}(y) = \tan ^{-1}(\frac{1}{y})$ for $y > 0$:
$\tan ^{-1}(\frac{1}{7})+\tan ^{-1}(\frac{1}{8})+\tan ^{-1}(\frac{1}{18})=\tan ^{-1}(\frac{1}{x})$
First,apply $\tan ^{-1}(A) + \tan ^{-1}(B) = \tan ^{-1}(\frac{A+B}{1-AB})$ for $\tan ^{-1}(\frac{1}{7}) + \tan ^{-1}(\frac{1}{8})$:
$\tan ^{-1}(\frac{\frac{1}{7}+\frac{1}{8}}{1-(\frac{1}{7})(\frac{1}{8})}) = \tan ^{-1}(\frac{\frac{15}{56}}{\frac{55}{56}}) = \tan ^{-1}(\frac{15}{55}) = \tan ^{-1}(\frac{3}{11})$
Now,add the third term:
$\tan ^{-1}(\frac{3}{11}) + \tan ^{-1}(\frac{1}{18}) = \tan ^{-1}(\frac{\frac{3}{11}+\frac{1}{18}}{1-(\frac{3}{11})(\frac{1}{18})})$
$= \tan ^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = \tan ^{-1}(\frac{65}{195}) = \tan ^{-1}(\frac{1}{3})$
Thus,$\tan ^{-1}(\frac{1}{3}) = \tan ^{-1}(\frac{1}{x})$,which implies $x = 3$.

(B) We know the following properties of inverse trigonometric functions: $\sec^{-1}(-x) = \pi - \sec^{-1}(x)$,$\operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1}(x)$,and $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$.
Also,$\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
$\sec^{-1}(-2) = \pi - \sec^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
$\operatorname{cosec}^{-1}(-\sqrt{2}) = -\operatorname{cosec}^{-1}(\sqrt{2}) = -\frac{\pi}{4}$.
$\cos^{-1}\left(\frac{-1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
Substituting these values into the expression:
$\frac{\frac{\pi}{3} - \frac{2\pi}{3}}{-\frac{\pi}{4} + \frac{2\pi}{3}} = \frac{-\frac{\pi}{3}}{\frac{-3\pi + 8\pi}{12}} = \frac{-\frac{\pi}{3}}{\frac{5\pi}{12}} = -\frac{\pi}{3} \times \frac{12}{5\pi} = -\frac{4}{5}$.

(C) We have $\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}$.
Using $\cos(\pi - \theta) = -\cos \theta$ and $\sin(\pi - \theta) = \sin \theta$,we get:
$= \cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(-\cos \frac{\pi}{10} - \sin \frac{\pi}{10}\right)\right\}$
$= \cos ^{-1}\left\{-\left(\frac{1}{\sqrt{2}} \cos \frac{\pi}{10} + \frac{1}{\sqrt{2}} \sin \frac{\pi}{10}\right)\right\}$
$= \cos ^{-1}\left\{-\left(\cos \frac{\pi}{4} \cos \frac{\pi}{10} + \sin \frac{\pi}{4} \sin \frac{\pi}{10}\right)\right\}$
$= \cos ^{-1}\left\{-\cos \left(\frac{\pi}{4} - \frac{\pi}{10}\right)\right\}$
$= \cos ^{-1}\left\{-\cos \left(\frac{5\pi - 2\pi}{20}\right)\right\} = \cos ^{-1}\left\{-\cos \frac{3\pi}{20}\right\}$
Since $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,we have:
$= \pi - \cos^{-1}\left(\cos \frac{3\pi}{20}\right)$
$= \pi - \frac{3\pi}{20} = \frac{17\pi}{20}$.

(D) Given,$\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi$
$\Rightarrow \cos ^{-1} x+\cos ^{-1} y=\pi-\cos ^{-1} z$
$\Rightarrow \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}(-z)$
$\Rightarrow \cos ^{-1}\left[x y-\sqrt{1-x^2} \sqrt{1-y^2}\right]=\cos ^{-1}(-z)$
$\Rightarrow x y-\sqrt{(1-x^2)(1-y^2)}=-z$
$\Rightarrow x y+z=\sqrt{(1-x^2)(1-y^2)}$
Squaring both sides:
$(x y+z)^2 = (1-x^2)(1-y^2)$
$x^2 y^2+2 x y z+z^2=1-y^2-x^2+x^2 y^2$
$x^2+y^2+z^2+2 x y z=1$
Comparing this with the given equation $x^2+y^2+z^2+k x y z=1$,we get $k=2$.

(C) Given equation: $\tan ^{-1}(x+2)+\tan ^{-1}(x-2) = \tan ^{-1}\left(\frac{1}{2}\right)$
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$\tan ^{-1}\left(\frac{(x+2)+(x-2)}{1-(x+2)(x-2)}\right) = \tan ^{-1}\left(\frac{1}{2}\right)$
$\tan ^{-1}\left(\frac{2x}{1-(x^2-4)}\right) = \tan ^{-1}\left(\frac{1}{2}\right)$
$\tan ^{-1}\left(\frac{2x}{5-x^2}\right) = \tan ^{-1}\left(\frac{1}{2}\right)$
Comparing both sides:
$\frac{2x}{5-x^2} = \frac{1}{2}$
$4x = 5 - x^2$
$x^2 + 4x - 5 = 0$
$(x+5)(x-1) = 0$
Thus,$x = -5$ or $x = 1$.

(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$
We can rewrite the expression as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)\right)$
Using the property $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get:
$\cos \left(\cos ^{-1} x + \frac{\pi}{2}\right)$
Using the trigonometric identity $\cos \left(\frac{\pi}{2} + \theta\right) = -\sin \theta$,we get:
$-\sin \left(\cos ^{-1} x\right)$
Since $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$,the expression becomes:
$-\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) = -\sqrt{1-x^2}$
Substituting $x = \frac{1}{5}$:
$-\sqrt{1-\left(\frac{1}{5}\right)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{\sqrt{24}}{5} = -\frac{2 \sqrt{6}}{5}$

(C) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$ and $\beta = \cos^{-1}\left(\frac{12}{13}\right)$.
Then $\sin \alpha = \frac{3}{5}$,which implies $\cos \alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$.
And $\cos \beta = \frac{12}{13}$,which implies $\sin \beta = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13}$.
Using the identity $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$:
$\cos(\alpha + \beta) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) - \left(\frac{3}{5}\right) \left(\frac{5}{13}\right)$
$= \frac{48}{65} - \frac{15}{65}$
$= \frac{33}{65}$.

(B) Let $x = \cos 2\theta$.
Then $\theta = \frac{1}{2} \cos^{-1} x$.
Substitute $x = \cos 2\theta$ into the expression:
$\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right)$
$= \tan^{-1}\left(\frac{\sqrt{2\cos^2\theta}-\sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta}+\sqrt{2\sin^2\theta}}\right)$
$= \tan^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right)$
$= \tan^{-1}\left(\frac{1 - \tan\theta}{1 + \tan\theta}\right)$
$= \tan^{-1}(\tan(\frac{\pi}{4} - \theta))$
$= \frac{\pi}{4} - \theta = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$.
Now,adding the second term:
$(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x) + \frac{1}{2} \cos^{-1} x = \frac{\pi}{4}$.

(A) We know that $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$ and $\sin ^{-1}(x) + \cos ^{-1}(x) = \frac{\pi}{2}$.
Let the given expression be $E = \sin \left(\cos ^{-1}\left(-\frac{1}{3}\right)-\sin ^{-1}\left(\frac{1}{3}\right)\right)$.
Using the property $\cos ^{-1}\left(-\frac{1}{3}\right) = \pi - \cos ^{-1}\left(\frac{1}{3}\right)$,we get:
$E = \sin \left(\pi - \cos ^{-1}\left(\frac{1}{3}\right) - \sin ^{-1}\left(\frac{1}{3}\right)\right)$.
Now,group the inverse trigonometric terms:
$E = \sin \left(\pi - \left(\cos ^{-1}\left(\frac{1}{3}\right) + \sin ^{-1}\left(\frac{1}{3}\right)\right)\right)$.
Since $\cos ^{-1}(x) + \sin ^{-1}(x) = \frac{\pi}{2}$ for $x \in [-1, 1]$,we have:
$E = \sin \left(\pi - \frac{\pi}{2}\right) = \sin \left(\frac{\pi}{2}\right)$.
Thus,$E = 1$.

(C) We use the formula $\tan ^{-1}(A) + \tan ^{-1}(B) = \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$.
Given $\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)$.
Applying the formula,we get $\tan ^{-1}\left[\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\left(\frac{x+1}{x-1}\right)\left(\frac{x-1}{x}\right)}\right]=\tan ^{-1}(-7)$.
Simplifying the expression inside the bracket:
$\frac{x(x+1)+(x-1)^2}{x(x-1)-(x+1)(x-1)} = \frac{x^2+x+x^2-2x+1}{x^2-x-(x^2-1)} = \frac{2x^2-x+1}{1-x}$.
Equating the arguments,we have $\frac{2x^2-x+1}{1-x} = -7$.
$2x^2-x+1 = -7(1-x) = -7+7x$.
$2x^2-8x+8 = 0$.
$x^2-4x+4 = 0$.
$(x-2)^2 = 0$,which gives $x=2$.

(B) We are given the equation $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} P$.
First,convert $\cos ^{-1}\left(\frac{12}{13}\right)$ into $\sin ^{-1}$ form.
Let $\cos ^{-1}\left(\frac{12}{13}\right) = \theta$,then $\cos \theta = \frac{12}{13}$.
Using $\sin \theta = \sqrt{1-\cos^2 \theta} = \sqrt{1-\left(\frac{12}{13}\right)^2} = \sqrt{1-\frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.
So,$\cos ^{-1}\left(\frac{12}{13}\right) = \sin ^{-1}\left(\frac{5}{13}\right)$.
Now,the equation becomes $\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right) = \sin ^{-1} P$.
Using the formula $\sin ^{-1} x + \sin ^{-1} y = \sin ^{-1}\left(x \sqrt{1-y^2} + y \sqrt{1-x^2}\right)$:
$\sin ^{-1} P = \sin ^{-1}\left[\frac{3}{5} \sqrt{1-\left(\frac{5}{13}\right)^2} + \frac{5}{13} \sqrt{1-\left(\frac{3}{5}\right)^2}\right]$
$= \sin ^{-1}\left[\frac{3}{5} \times \frac{12}{13} + \frac{5}{13} \times \frac{4}{5}\right]$
$= \sin ^{-1}\left(\frac{36}{65} + \frac{20}{65}\right) = \sin ^{-1}\left(\frac{56}{65}\right)$.
Therefore,$P = \frac{56}{65}$.

(C) We use the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left( \frac{a+b}{1-ab} \right)$.
$\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4}\right)\left(\frac{2}{9}\right)}\right)$
$= \tan ^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right) = \tan ^{-1}\left(\frac{17}{34}\right) = \tan ^{-1}\left(\frac{1}{2}\right)$
Now,we use the identity $2 \tan ^{-1} \theta = \cos ^{-1} \left( \frac{1-\theta^2}{1+\theta^2} \right)$,which implies $\tan ^{-1} \theta = \frac{1}{2} \cos ^{-1} \left( \frac{1-\theta^2}{1+\theta^2} \right)$.
Setting $\theta = \frac{1}{2}$,we get $\tan ^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} \cos ^{-1} \left( \frac{1-(1/2)^2}{1+(1/2)^2} \right)$
$= \frac{1}{2} \cos ^{-1} \left( \frac{1-1/4}{1+1/4} \right) = \frac{1}{2} \cos ^{-1} \left( \frac{3/4}{5/4} \right) = \frac{1}{2} \cos ^{-1} \left( \frac{3}{5} \right)$
Comparing this with $\frac{1}{2} \cos ^{-1} x$,we find $x = \frac{3}{5}$.

(D) We know that $\tan ^{-1}(1) = \frac{\pi}{4}$.
We also know that $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$,so $\cos ^{-1}\left(-\frac{1}{2}\right) = \pi - \cos ^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
We know that $\sin ^{-1}(-x) = -\sin ^{-1}(x)$,so $\sin ^{-1}\left(-\frac{1}{2}\right) = -\sin ^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}$.
Adding these values together:
$\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$.

(C) We use the formula $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$ for $|x| < 1$.
Applying this to $2 \tan^{-1} \frac{1}{2}$:
$2 \tan^{-1} \frac{1}{2} = \tan^{-1} \left( \frac{2(1/2)}{1-(1/2)^2} \right) = \tan^{-1} \left( \frac{1}{1-1/4} \right) = \tan^{-1} \left( \frac{1}{3/4} \right) = \tan^{-1} \left( \frac{4}{3} \right)$.
Now,add $\tan^{-1} \frac{1}{7}$ using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$:
$\tan^{-1} \left( \frac{4}{3} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \tan^{-1} \left( \frac{4/3 + 1/7}{1 - (4/3)(1/7)} \right)$.
Calculate the numerator: $\frac{4}{3} + \frac{1}{7} = \frac{28+3}{21} = \frac{31}{21}$.
Calculate the denominator: $1 - \frac{4}{21} = \frac{21-4}{21} = \frac{17}{21}$.
Thus,the expression becomes $\tan^{-1} \left( \frac{31/21}{17/21} \right) = \tan^{-1} \left( \frac{31}{17} \right)$.

(A) Given equation: $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$
Let $t=\cos ^{-1} \sqrt{p}$. Then $\sqrt{p}=\cos t$,so $p=\cos ^2 t$.
Thus,$1-p=1-\cos ^2 t=\sin ^2 t$,which implies $\sqrt{1-p}=\sin t$.
Therefore,$t=\sin ^{-1} \sqrt{1-p}$.
Since $t=\cos ^{-1} \sqrt{p}$,we have $\cos ^{-1} \sqrt{p}=\sin ^{-1} \sqrt{1-p}$.
Substituting this into the given equation:
$\sin ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$
Using the identity $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$,we get:
$\frac{\pi}{2}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$
$\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}-\frac{\pi}{2}=\frac{\pi}{4}$
$\sqrt{1-q}=\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$
Squaring both sides: $1-q=\frac{1}{2}$
$q=1-\frac{1}{2}=\frac{1}{2}$

(B) We use the formula $\tan ^{-1}(x) + \tan ^{-1}(y) = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$.
First,calculate $\tan ^{-1}\left(\frac{1}{2}\right) + \tan ^{-1}\left(\frac{1}{5}\right)$:
$\tan ^{-1}\left(\frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \cdot \frac{1}{5}}\right) = \tan ^{-1}\left(\frac{\frac{7}{10}}{\frac{9}{10}}\right) = \tan ^{-1}\left(\frac{7}{9}\right)$.
Now,add $\tan ^{-1}\left(\frac{1}{8}\right)$ to the result:
$\tan ^{-1}\left(\frac{1}{8}\right) + \tan ^{-1}\left(\frac{7}{9}\right) = \tan ^{-1}\left(\frac{\frac{1}{8} + \frac{7}{9}}{1 - \frac{1}{8} \cdot \frac{7}{9}}\right)$.
Simplify the expression inside the bracket:
$\frac{\frac{9+56}{72}}{\frac{72-7}{72}} = \frac{65}{65} = 1$.
Thus,the expression becomes $\tan ^{-1}(1) = \frac{\pi}{4}$.

(A) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,group the terms:
$\left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} \right) + \left( \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8} \right)$
Apply the formula to the first pair:
$\tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$
Apply the formula to the second pair:
$\tan ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) = \tan ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$
Now add the results:
$\tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \times \frac{3}{11}} \right)$
$= \tan ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} \right) = \tan ^{-1} \left( \frac{65}{65} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$

(D) We use the formula $2 \tan ^{-1}(x) = \tan ^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$.
First,calculate $2 \tan ^{-1}\left(\frac{1}{3}\right) = \tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-(\frac{1}{3})^2}\right) = \tan ^{-1}\left(\frac{2/3}{8/9}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$.
Now,convert $\cos ^{-1}\left(\frac{3}{5}\right)$ to $\tan ^{-1}$. Let $\theta = \cos ^{-1}\left(\frac{3}{5}\right)$,then $\cos \theta = \frac{3}{5}$. Since $\cos^2 \theta + \sin^2 \theta = 1$,$\sin \theta = \sqrt{1 - (3/5)^2} = 4/5$. Thus,$\tan \theta = \frac{4/5}{3/5} = \frac{4}{3}$,so $\cos ^{-1}\left(\frac{3}{5}\right) = \tan ^{-1}\left(\frac{4}{3}\right)$.
The expression becomes $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{4}{3}\right)$.
Using $\tan ^{-1}(x) + \tan ^{-1}(1/x) = \frac{\pi}{2}$ for $x > 0$,we get $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{1}{3/4}\right) = \frac{\pi}{2}$.

(B) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,group the terms:
$\left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} \right) + \left( \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8} \right)$
Applying the formula to the first pair:
$\tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$
Applying the formula to the second pair:
$\tan ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) = \tan ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$
Now,add the results:
$\tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \times \frac{3}{11}} \right)$
$= \tan ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} \right) = \tan ^{-1} \left( \frac{65}{65} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$

(D) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \pi + \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ when $xy > 1$.
Here,$x = 2$ and $y = 3$. Since $xy = 2 \times 3 = 6 > 1$,we apply the formula:
$\tan ^{-1} 2 + \tan ^{-1} 3 = \pi + \tan ^{-1} \left( \frac{2+3}{1-(2 \times 3)} \right)$
$= \pi + \tan ^{-1} \left( \frac{5}{1-6} \right)$
$= \pi + \tan ^{-1} \left( \frac{5}{-5} \right)$
$= \pi + \tan ^{-1} (-1)$
Since $\tan ^{-1} (-1) = -\frac{\pi}{4}$,we have:
$= \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.

(B) We know that $2 \tan ^{-1}(x) = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
Therefore,$2 \tan ^{-1} \left( \frac{1}{5} \right) = \tan ^{-1} \left( \frac{2 \times \frac{1}{5}}{1 - (\frac{1}{5})^2} \right) = \tan ^{-1} \left( \frac{\frac{2}{5}}{1 - \frac{1}{25}} \right) = \tan ^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan ^{-1} \left( \frac{2}{5} \times \frac{25}{24} \right) = \tan ^{-1} \left( \frac{5}{12} \right)$.
Now,the expression becomes $\tan \left[ \tan ^{-1} \left( \frac{5}{12} \right) - \tan ^{-1} (1) \right]$.
Using the formula $\tan ^{-1}(x) - \tan ^{-1}(y) = \tan ^{-1} \left( \frac{x-y}{1+xy} \right)$,we get:
$\tan ^{-1} \left( \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \times 1} \right) = \tan ^{-1} \left( \frac{-\frac{7}{12}}{\frac{17}{12}} \right) = \tan ^{-1} \left( -\frac{7}{17} \right)$.
Finally,$\tan \left[ \tan ^{-1} \left( -\frac{7}{17} \right) \right] = -\frac{7}{17}$.

(D) We know that $\cos ^{-1}\left(\frac{4}{5}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$ because in a right-angled triangle with base $4$ and hypotenuse $5$,the perpendicular is $\sqrt{5^2 - 4^2} = 3$.
Substituting this into the expression:
$\tan \left(\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{2}{3}\right)\right)$
Using the formula $\tan ^{-1}(x) + \tan ^{-1}(y) = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$:
$= \tan \left(\tan ^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{17}{6}\right)\right) = \frac{17}{6}$

(B) Given the equation: $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$
Using the identity $\tan ^{-1}(A)+\tan ^{-1}(B)=\tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$\tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{5 x}{1-6 x^2}\right)=\frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{5 x}{1-6 x^2}=\tan\left(\frac{\pi}{4}\right)=1$
$5 x = 1 - 6 x^2$
$6 x^2 + 5 x - 1 = 0$
Factoring the quadratic equation:
$(6 x - 1)(x + 1) = 0$
This gives $x = \frac{1}{6}$ or $x = -1$.
Since the problem states $x > 0$,we discard $x = -1$.
Therefore,$x = \frac{1}{6}$.

(A) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \pi + \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ when $xy > 1$.
Here,$x = 2$ and $y = 3$,so $xy = 6 > 1$.
Therefore,$\tan ^{-1} 2 + \tan ^{-1} 3 = \pi + \tan ^{-1} \left( \frac{2+3}{1-(2)(3)} \right)$.
$= \pi + \tan ^{-1} \left( \frac{5}{1-6} \right) = \pi + \tan ^{-1} (-1)$.
$= \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.
Thus,the value is $\left( \frac{3 \pi}{4} \right)^c$.

(C) Given equation is $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$.
Rearranging the terms,we get $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$.
Using the property $\tan ^{-1} a - \tan ^{-1} b = \tan ^{-1}\left(\frac{a-b}{1+ab}\right)$,we can write $\tan ^{-1}(1) - \tan ^{-1}(x) = \frac{1}{2} \tan ^{-1} x$.
This simplifies to $\frac{\pi}{4} = \frac{1}{2} \tan ^{-1} x + \tan ^{-1} x$.
$\frac{\pi}{4} = \frac{3}{2} \tan ^{-1} x$.
$\tan ^{-1} x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.
Therefore,$x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

(A) We use the formula $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
For $x = \frac{1}{3}$,we have:
$2 \tan ^{-1} \left( \frac{1}{3} \right) = \tan ^{-1} \left( \frac{2 \times \frac{1}{3}}{1 - (\frac{1}{3})^2} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} \right)$
$= \tan ^{-1} \left( \frac{2}{3} \times \frac{9}{8} \right) = \tan ^{-1} \left( \frac{3}{4} \right)$
Now,substituting this back into the original expression:
$2 \tan ^{-1} \left( \frac{1}{3} \right) - \tan ^{-1} \left( \frac{3}{4} \right) = \tan ^{-1} \left( \frac{3}{4} \right) - \tan ^{-1} \left( \frac{3}{4} \right) = 0$.

(C) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,combine the first two terms:
$\tan ^{-1} \left( \frac{1}{3} \right) + \tan ^{-1} \left( \frac{1}{5} \right) = \tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$.
Now,combine the last two terms:
$\tan ^{-1} \left( \frac{1}{7} \right) + \tan ^{-1} \left( \frac{1}{8} \right) = \tan ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) = \tan ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$.
Finally,add the results:
$\tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \times \frac{3}{11}} \right) = \tan ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} \right) = \tan ^{-1} \left( \frac{65}{65} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$.
Thus,the correct option is $C$.

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

(D) We know that $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
Applying this for $x = \frac{1}{2}$,we get $2 \tan ^{-1} \frac{1}{2} = \tan ^{-1} \left( \frac{2 \times (1/2)}{1 - (1/2)^2} \right) = \tan ^{-1} \left( \frac{1}{1 - 1/4} \right) = \tan ^{-1} \left( \frac{1}{3/4} \right) = \tan ^{-1} \left( \frac{4}{3} \right)$.
Now,the expression becomes $\tan ^{-1} \left( \frac{4}{3} \right) + \tan ^{-1} \left( \frac{3}{8} \right)$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$,we have:
$\tan ^{-1} \left( \frac{4/3 + 3/8}{1 - (4/3)(3/8)} \right) = \tan ^{-1} \left( \frac{(32+9)/24}{1 - 12/24} \right) = \tan ^{-1} \left( \frac{41/24}{12/24} \right) = \tan ^{-1} \left( \frac{41}{12} \right)$.

(C) Let $x = \tan \theta$,where $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
Then the given expression becomes:
$3 \sin ^{-1}(\sin 2\theta) - 4 \cos ^{-1}(\cos 2\theta) + 2 \tan ^{-1}(\tan 2\theta) = \frac{\pi}{3}$.
Assuming $x \in (-1, 1)$,then $2\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
Thus,$3(2\theta) - 4(2\theta) + 2(2\theta) = \frac{\pi}{3}$.
$6\theta - 8\theta + 4\theta = \frac{\pi}{3}$.
$2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6}$.
Therefore,$x = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.

(C) Let $\theta = 2 \tan^{-1} \frac{1}{5}$.
Using the formula $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$,we have:
$\theta = \tan^{-1} \left( \frac{2(1/5)}{1-(1/5)^2} \right) = \tan^{-1} \left( \frac{2/5}{1-1/25} \right) = \tan^{-1} \left( \frac{2/5}{24/25} \right) = \tan^{-1} \left( \frac{2}{5} \times \frac{25}{24} \right) = \tan^{-1} \left( \frac{5}{12} \right)$.
Now,we need to evaluate $\tan \left( \theta - \frac{\pi}{4} \right)$.
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,where $A = \tan^{-1} \frac{5}{12}$ and $B = \frac{\pi}{4}$:
$\tan \left( \tan^{-1} \frac{5}{12} - \frac{\pi}{4} \right) = \frac{\frac{5}{12} - \tan \frac{\pi}{4}}{1 + \frac{5}{12} \tan \frac{\pi}{4}} = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12}(1)} = \frac{\frac{5-12}{12}}{\frac{12+5}{12}} = \frac{-7}{17}$.

(C) Let $\theta = \tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)$. Then $\tan \theta = \frac{\sqrt{5}-1}{2}$.
We need to find $\tan(2\theta)$.
Using the formula $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$:
$\tan(2\theta) = \frac{2 \left(\frac{\sqrt{5}-1}{2}\right)}{1 - \left(\frac{\sqrt{5}-1}{2}\right)^2}$
$= \frac{\sqrt{5}-1}{1 - \frac{5 + 1 - 2\sqrt{5}}{4}}$
$= \frac{\sqrt{5}-1}{1 - \frac{6 - 2\sqrt{5}}{4}}$
$= \frac{\sqrt{5}-1}{\frac{4 - 6 + 2\sqrt{5}}{4}}$
$= \frac{4(\sqrt{5}-1)}{2\sqrt{5}-2}$
$= \frac{4(\sqrt{5}-1)}{2(\sqrt{5}-1)}$
$= 2$.

(C) The general term of the series is $T_n = \cot^{-1}(2n^2)$.
We know that $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$ for $x > 0$.
So,$T_n = \tan^{-1}(\frac{1}{2n^2})$.
We can rewrite the argument as $\frac{1}{2n^2} = \frac{2}{4n^2} = \frac{2}{1 + (2n^2 - 1)} = \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$.
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}(\frac{x-y}{1+xy})$,we get:
$T_n = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$.
The sum of the first $N$ terms is $S_N = \sum_{n=1}^{N} [\tan^{-1}(2n+1) - \tan^{-1}(2n-1)]$.
This is a telescoping series:
$S_N = (\tan^{-1}(3) - \tan^{-1}(1)) + (\tan^{-1}(5) - \tan^{-1}(3)) + \ldots + (\tan^{-1}(2N+1) - \tan^{-1}(2N-1))$.
$S_N = \tan^{-1}(2N+1) - \tan^{-1}(1)$.
As $N \to \infty$,$S_N = \tan^{-1}(\infty) - \tan^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(B) We know that $\sum_{k=1}^n 2k = 2 \times \frac{n(n+1)}{2} = n(n+1)$.
Substituting this into the expression,we get $\cot \left(\sum_{n=1}^{23} \cot ^{-1}(1+n(n+1))\right)$.
Using the identity $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$ for $x > 0$,the expression becomes $\cot \left(\sum_{n=1}^{23} \tan ^{-1}\left(\frac{1}{1+n(n+1)}\right)\right)$.
We can rewrite the argument of $\tan^{-1}$ as $\frac{(n+1)-n}{1+n(n+1)}$,which allows us to use the formula $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a-b}{1+ab})$.
Thus,$\sum_{n=1}^{23} \tan ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) = \sum_{n=1}^{23} (\tan^{-1}(n+1) - \tan^{-1}(n))$.
This is a telescoping sum: $(\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + \dots + (\tan^{-1}(24) - \tan^{-1}(23)) = \tan^{-1}(24) - \tan^{-1}(1)$.
Now,$\cot(\tan^{-1}(24) - \tan^{-1}(1)) = \cot(\tan^{-1}(\frac{24-1}{1+24 \times 1})) = \cot(\tan^{-1}(\frac{23}{25}))$.
Since $\cot(\tan^{-1}(x)) = \cot(\cot^{-1}(\frac{1}{x})) = \frac{1}{x}$,we have $\cot(\tan^{-1}(\frac{23}{25})) = \frac{25}{23}$.