Properties of ITF Questions in English

(B) Given $\sec^{-1} \left( \frac{a+b}{a-b} \right) = 2 \sin^{-1} x$.
Taking the reciprocal,we get $\cos^{-1} \left( \frac{a-b}{a+b} \right) = 2 \sin^{-1} x$.
Dividing the numerator and denominator by $a$,we have $\cos^{-1} \left( \frac{1 - b/a}{1 + b/a} \right) = 2 \sin^{-1} x$.
Using the identity $\cos^{-1} \left( \frac{1 - y^2}{1 + y^2} \right) = 2 \tan^{-1} y$,where $y = \sqrt{b/a}$,we get $2 \tan^{-1} \left( \sqrt{\frac{b}{a}} \right) = 2 \sin^{-1} x$.
Dividing by $2$,we have $\tan^{-1} \left( \sqrt{\frac{b}{a}} \right) = \sin^{-1} x$.
Using the identity $\tan^{-1} y = \sin^{-1} \left( \frac{y}{\sqrt{1+y^2}} \right)$,we get $\sin^{-1} \left( \frac{\sqrt{b/a}}{\sqrt{1 + b/a}} \right) = \sin^{-1} x$.
Simplifying the expression inside,we get $\sin^{-1} \left( \sqrt{\frac{b/a}{(a+b)/a}} \right) = \sin^{-1} x$.
Thus,$x = \sqrt{\frac{b}{a+b}}$.

(A) Let $\sin^{-1} x = y$. Then,$x = \sin y$.
Since $-1 \leq x < 0$,we have $-\frac{\pi}{2} \leq \sin^{-1} x < 0$,which implies $-\frac{\pi}{2} \leq y < 0$.
We know that $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.
Since $y$ is in the interval $[-\frac{\pi}{2}, 0)$,$-y$ is in the interval $(0, \frac{\pi}{2}]$.
Using the property $\cos(-y) = \cos y = \sqrt{1 - x^2}$,we get $-y = \cos^{-1}(\sqrt{1 - x^2})$.
Therefore,$y = -\cos^{-1}(\sqrt{1 - x^2})$.
Thus,$\sin^{-1} x = -\cos^{-1} \sqrt{1 - x^2}$.

(A) Given the inequality: $ a + \frac{\pi}{2} < 2 \tan^{-1} x + 3 \cot^{-1} x < b $.
We know that $ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} $,so $ \tan^{-1} x = \frac{\pi}{2} - \cot^{-1} x $.
Substituting this into the expression: $ 2(\frac{\pi}{2} - \cot^{-1} x) + 3 \cot^{-1} x = \pi - 2 \cot^{-1} x + 3 \cot^{-1} x = \pi + \cot^{-1} x $.
Now the inequality becomes: $ a + \frac{\pi}{2} < \pi + \cot^{-1} x < b $.
Subtracting $ \pi $ from all parts: $ a - \frac{\pi}{2} < \cot^{-1} x < b - \pi $.
Since the range of $ \cot^{-1} x $ is $ (0, \pi) $,we have $ 0 < \cot^{-1} x < \pi $.
Comparing $ a - \frac{\pi}{2} < \cot^{-1} x < b - \pi $ with $ 0 < \cot^{-1} x < \pi $,we get:
$ a - \frac{\pi}{2} = 0 \Rightarrow a = \frac{\pi}{2} $.
$ b - \pi = \pi \Rightarrow b = 2\pi $.
Thus,$ a = \frac{\pi}{2} $ and $ b = 2\pi $.

(B) Given that,$\sin^{-1} x + \cos^{-1} y = \frac{2\pi}{5} \quad (1)$
We know the fundamental identities: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $\sin^{-1} y + \cos^{-1} y = \frac{\pi}{2}$.
From these,we can write $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$ and $\cos^{-1} y = \frac{\pi}{2} - \sin^{-1} y$.
Substituting these into equation $(1)$:
$(\frac{\pi}{2} - \cos^{-1} x) + (\frac{\pi}{2} - \sin^{-1} y) = \frac{2\pi}{5}$
$\pi - (\cos^{-1} x + \sin^{-1} y) = \frac{2\pi}{5}$
$\cos^{-1} x + \sin^{-1} y = \pi - \frac{2\pi}{5}$
$\cos^{-1} x + \sin^{-1} y = \frac{3\pi}{5}$

(B) Given the equation: $3 \tan^{-1} x + \cot^{-1} x = \pi$.
We know that $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
We can rewrite the given equation as: $2 \tan^{-1} x + (\tan^{-1} x + \cot^{-1} x) = \pi$.
Substituting the identity $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ into the equation:
$2 \tan^{-1} x + \frac{\pi}{2} = \pi$.
Subtracting $\frac{\pi}{2}$ from both sides:
$2 \tan^{-1} x = \frac{\pi}{2}$.
Dividing by $2$:
$\tan^{-1} x = \frac{\pi}{4}$.
Taking the tangent of both sides:
$x = \tan(\frac{\pi}{4}) = 1$.

(A) Given that,$\sin ^{-1} x + \sin ^{-1} y = \frac{\pi}{2} \dots (1)$
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2} \dots (2)$
Comparing equations $(1)$ and $(2)$,we get $\sin ^{-1} y = \cos ^{-1} x$
Taking $\cos$ on both sides,we have $\cos(\sin ^{-1} y) = \cos(\cos ^{-1} x)$
Since $\cos(\sin ^{-1} y) = \sqrt{1 - y^{2}}$,we get $x = \sqrt{1 - y^{2}}$
Squaring both sides,we obtain $x^{2} = 1 - y^{2}$

(A) Given expression: $\frac{\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)}{\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)}$
We know that $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in [-1, 1]$ and $\tan ^{-1} \theta + \cot ^{-1} \theta = \frac{\pi}{2}$ for $\theta \in \mathbb{R}$.
Using the identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $\sin x = \cos(\frac{\pi}{2} - x)$:
Numerator: $\sin ^{-1}(\sin(\frac{\pi}{2} - x)) + \cos ^{-1}(\cos(\frac{\pi}{2} - x)) = (\frac{\pi}{2} - x) + (\frac{\pi}{2} - x) = \pi - 2x$.
Denominator: $\tan ^{-1}(\tan(\frac{\pi}{2} - x)) + \cot ^{-1}(\cot(\frac{\pi}{2} - x)) = (\frac{\pi}{2} - x) + (\frac{\pi}{2} - x) = \pi - 2x$.
Thus,the expression becomes $\frac{\pi - 2x}{\pi - 2x} = 1$.

(B) Given that $2 \sin ^{-1} x - 3 \cos ^{-1} x = 4$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,so $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the given equation:
$2 \sin ^{-1} x - 3(\frac{\pi}{2} - \sin ^{-1} x) = 4$
$2 \sin ^{-1} x - \frac{3 \pi}{2} + 3 \sin ^{-1} x = 4$
$5 \sin ^{-1} x = 4 + \frac{3 \pi}{2}$
$\sin ^{-1} x = \frac{8 + 3 \pi}{10}$
Now,we need to find the value of $2 \sin ^{-1} x + 3 \cos ^{-1} x$.
$2 \sin ^{-1} x + 3 \cos ^{-1} x = 2 \sin ^{-1} x + 3(\frac{\pi}{2} - \sin ^{-1} x)$
$= 2 \sin ^{-1} x + \frac{3 \pi}{2} - 3 \sin ^{-1} x$
$= \frac{3 \pi}{2} - \sin ^{-1} x$
$= \frac{3 \pi}{2} - \frac{8 + 3 \pi}{10}$
$= \frac{15 \pi - 8 - 3 \pi}{10}$
$= \frac{12 \pi - 8}{10} = \frac{6 \pi - 4}{5}$

(B) 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 $x$:
$\tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{1 - \frac{y}{x}}{1 + \frac{y}{x}}\right)$.
Using the identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$,we have:
$\tan^{-1}\left(\frac{1 - \frac{y}{x}}{1 + 1 \cdot \frac{y}{x}}\right) = \tan^{-1}(1) - \tan^{-1}\left(\frac{y}{x}\right)$.
Substituting this back into the expression:
$= \tan^{-1}\left(\frac{x}{y}\right) - \left(\tan^{-1}(1) - \tan^{-1}\left(\frac{y}{x}\right)\right)$
$= \tan^{-1}\left(\frac{x}{y}\right) + \tan^{-1}\left(\frac{y}{x}\right) - \tan^{-1}(1)$.
Since $\tan^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$ and $\cot^{-1}(z) = \tan^{-1}\left(\frac{1}{z}\right)$:
$= \tan^{-1}\left(\frac{x}{y}\right) + \cot^{-1}\left(\frac{x}{y}\right) - \tan^{-1}(1)$
$= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(A) We use the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left( \frac{A-B}{1+AB} \right)$.
Let $A = \frac{x}{y}$ and $B = \frac{x-y}{x+y}$.
Then,$\tan ^{-1} \left( \frac{x}{y} \right) - \tan ^{-1} \left( \frac{x-y}{x+y} \right) = \tan ^{-1} \left( \frac{\frac{x}{y} - \frac{x-y}{x+y}}{1 + \frac{x}{y} \cdot \frac{x-y}{x+y}} \right)$.
Simplifying the numerator: $\frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2 + xy - xy + y^2}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$.
Simplifying the denominator: $1 + \frac{x^2 - xy}{y(x+y)} = \frac{xy + y^2 + x^2 - xy}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$.
Thus,the expression becomes $\tan ^{-1} \left( \frac{\frac{x^2 + y^2}{y(x+y)}}{\frac{x^2 + y^2}{y(x+y)}} \right) = \tan ^{-1}(1)$.
Since $\tan ^{-1}(1) = \frac{\pi}{4}$,the final value is $\frac{\pi}{4}$.

(A) Let $\theta = \sin^{-1} \sqrt{\frac{63}{65}}$. Then $\sin \theta = \sqrt{\frac{63}{65}}$.
We need to find $\sin(2\theta)$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$.
First,find $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{63}{65}} = \sqrt{\frac{2}{65}}$.
Now,substitute the values into the identity:
$\sin(2\theta) = 2 \times \sqrt{\frac{63}{65}} \times \sqrt{\frac{2}{65}}$
$\sin(2\theta) = 2 \times \frac{\sqrt{63 \times 2}}{65} = 2 \times \frac{\sqrt{126}}{65} = \frac{2 \sqrt{126}}{65}$.

(A) Let $y = \cos ^{-1}(2x^2 - 1)$ and $z = \cos ^{-1} x$.
We know that $\cos ^{-1}(2x^2 - 1) = 2 \cos ^{-1} x$ for $x \in [0, 1]$.
Therefore,$y = 2z$.
Differentiating $y$ with respect to $z$,we get $\frac{dy}{dz} = \frac{d}{dz}(2z) = 2$.

(C) Let the given expression be $E = \tan \left[\sin ^{-1}\left\{\frac{x}{\sqrt{2}}+\frac{\sqrt{1-x^{2}}}{\sqrt{2}}\right\}-\sin ^{-1} x\right]$.
Substitute $x = \sin \theta$,where $\theta = \sin^{-1} x$.
Since $0 \leq x \leq \frac{1}{2}$,we have $0 \leq \theta \leq \frac{\pi}{6}$.
The expression inside the $\sin^{-1}$ becomes $\frac{\sin \theta}{\sqrt{2}} + \frac{\cos \theta}{\sqrt{2}} = \sin \theta \cos \frac{\pi}{4} + \cos \theta \sin \frac{\pi}{4} = \sin \left(\theta + \frac{\pi}{4}\right)$.
Now,the expression becomes $E = \tan \left[\sin ^{-1}\left\{\sin \left(\theta + \frac{\pi}{4}\right)\right\} - \theta\right]$.
Since $0 \leq \theta \leq \frac{\pi}{6}$,then $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12}$.
Since $\frac{\pi}{4} \leq \theta + \frac{\pi}{4} \leq \frac{5\pi}{12}$,the value lies within the principal range of $\sin^{-1}$,so $\sin^{-1}(\sin(\theta + \frac{\pi}{4})) = \theta + \frac{\pi}{4}$.
Thus,$E = \tan \left(\theta + \frac{\pi}{4} - \theta\right) = \tan \frac{\pi}{4} = 1$.

(A) We are given that $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$.
We know the identity $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ for any real $x$.
This implies $\tan ^{-1} x=\frac{\pi}{2}-\cot ^{-1} x$ and $\tan ^{-1} y=\frac{\pi}{2}-\cot ^{-1} y$.
Substituting these into the given equation:
$(\frac{\pi}{2}-\cot ^{-1} x) + (\frac{\pi}{2}-\cot ^{-1} y) = \frac{4 \pi}{5}$.
$\pi - (\cot ^{-1} x + \cot ^{-1} y) = \frac{4 \pi}{5}$.
$\cot ^{-1} x + \cot ^{-1} y = \pi - \frac{4 \pi}{5}$.
$\cot ^{-1} x + \cot ^{-1} y = \frac{\pi}{5}$.

(D) Given,$\frac{(x+1)^{2}}{x^{3}+x} = \frac{A}{x} + \frac{Bx+C}{x^{2}+1} \dots (i)$
Multiplying both sides by $x(x^{2}+1)$,we get:
$(x+1)^{2} = A(x^{2}+1) + (Bx+C)x$
$x^{2} + 2x + 1 = Ax^{2} + A + Bx^{2} + Cx$
$x^{2} + 2x + 1 = (A+B)x^{2} + Cx + A$
Comparing the coefficients of $x^{2}$,$x$,and the constant term on both sides:
$A+B = 1$
$C = 2$
$A = 1$
Substituting $A=1$ into $A+B=1$,we get $1+B=1$,so $B=0$.
Now,we need to calculate $\sin^{-1} A + \tan^{-1} B + \sec^{-1} C$:
$\sin^{-1}(1) + \tan^{-1}(0) + \sec^{-1}(2)$
$= \frac{\pi}{2} + 0 + \frac{\pi}{3}$
$= \frac{3\pi + 2\pi}{6} = \frac{5\pi}{6}$

(C) Given the expression $f(x) = 2 \sin^{-1} x + \cos^{-1} x$.
We can rewrite this as $f(x) = \sin^{-1} x + (\sin^{-1} x + \cos^{-1} x)$.
Since we know the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression becomes $f(x) = \sin^{-1} x + \frac{\pi}{2}$.
The range of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Adding $\frac{\pi}{2}$ to all parts of the inequality,we get $-\frac{\pi}{2} + \frac{\pi}{2} \leq \sin^{-1} x + \frac{\pi}{2} \leq \frac{\pi}{2} + \frac{\pi}{2}$.
This simplifies to $0 \leq 2 \sin^{-1} x + \cos^{-1} x \leq \pi$.
Comparing this with $\alpha \leq 2 \sin^{-1} x + \cos^{-1} x \leq \beta$,we find $\alpha = 0$ and $\beta = \pi$.

(B) We know that for any $x \in [-1, 1]$,the identity $\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}$ holds true.
Given the expression $\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$.
Substituting $x = -\frac{1}{7}$ into the identity,we get:
$\cos \left[\frac{\pi}{2}\right]$
Since $\cos \frac{\pi}{2} = 0$,the value of the expression is $0$.

(D) Given: $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$
Taking $\cos$ on both sides:
$\cos(\cos ^{-1} 2x + \cos ^{-1} 3x) = \cos(\frac{\pi}{3})$
Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$:
$(2x)(3x) - \sqrt{1-(2x)^2} \sqrt{1-(3x)^2} = \frac{1}{2}$
$6x^2 - \frac{1}{2} = \sqrt{1-4x^2} \sqrt{1-9x^2}$
Squaring both sides:
$(6x^2 - \frac{1}{2})^2 = (1-4x^2)(1-9x^2)$
$36x^4 - 6x^2 + \frac{1}{4} = 1 - 13x^2 + 36x^4$
$7x^2 = \frac{3}{4}$
$x^2 = \frac{3}{28}$
Then $4x^2 = 4 \times \frac{3}{28} = \frac{3}{7} = \frac{a}{b}$
Thus,$a = 3$ and $b = 7$.
$a + b = 3 + 7 = 10$.

(C) Given the equation: $\sin \left(2 \tan ^{-1} \frac{3}{4}\right)=\cos \left(2 \tan ^{-1} x\right)$
Using the formula $\sin(2 \tan^{-1} \theta) = \frac{2\theta}{1+\theta^2}$ and $\cos(2 \tan^{-1} x) = \frac{1-x^2}{1+x^2}$:
$\frac{2 \times \frac{3}{4}}{1+\left(\frac{3}{4}\right)^2} = \frac{1-x^2}{1+x^2}$
$\frac{\frac{3}{2}}{1+\frac{9}{16}} = \frac{1-x^2}{1+x^2}$
$\frac{\frac{3}{2}}{\frac{25}{16}} = \frac{1-x^2}{1+x^2}$
$\frac{3}{2} \times \frac{16}{25} = \frac{24}{25} = \frac{1-x^2}{1+x^2}$
$24(1+x^2) = 25(1-x^2)$
$24 + 24x^2 = 25 - 25x^2$
$49x^2 = 1$
$x^2 = \frac{1}{49}$
$x = \frac{1}{7}$

(B) Let $x = \sinh^{-1}(\sqrt{8})$ and $y = \cosh^{-1}(5)$.
Then $\sinh x = \sqrt{8}$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = 3$.
Thus $e^x = \cosh x + \sinh x = 3 + \sqrt{8} = 3 + 2\sqrt{2}$.
For $y = \cosh^{-1}(5)$,$\cosh y = 5$ and $\sinh y = \sqrt{\cosh^2 y - 1} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.
Thus $e^y = \cosh y + \sinh y = 5 + 2\sqrt{6}$.
We need to evaluate $\cosh(x + y) = \frac{e^{x+y} + e^{-(x+y)}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$.
Note that $e^{-x} = \frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}$ and $e^{-y} = \frac{1}{5 + 2\sqrt{6}} = 5 - 2\sqrt{6}$.
$e^x e^y = (3 + 2\sqrt{2})(5 + 2\sqrt{6}) = 15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}$.
$e^{-x} e^{-y} = (3 - 2\sqrt{2})(5 - 2\sqrt{6}) = 15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3}$.
$\cosh(x + y) = \frac{(15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}) + (15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3})}{2} = \frac{30 + 16\sqrt{3}}{2} = 15 + 8\sqrt{3}$.

(B) We are given the limit: $\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$.
Since $x \rightarrow 0^{+}$,we use the following trigonometric identities:
$2 \tan ^{-1} x = \sin ^{-1} \left(\frac{2 x}{1+x^2}\right) = \cos ^{-1} \left(\frac{1-x^2}{1+x^2}\right)$
$3 \tan ^{-1} x = \tan ^{-1} \left(\frac{3 x-x^3}{1-3 x^2}\right)$
Substituting these into the limit expression:
$\lim _{x \rightarrow 0^{+}} \frac{x \cdot (2 \tan ^{-1} x)}{(2 \tan ^{-1} x) \cdot (3 \tan ^{-1} x)} = \lim _{x \rightarrow 0^{+}} \frac{x}{3 \tan ^{-1} x}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x} = 1$,we get:
$\lim _{x \rightarrow 0^{+}} \frac{1}{3 \left(\frac{\tan ^{-1} x}{x}\right)} = \frac{1}{3 \times 1} = \frac{1}{3}$.

(C) Given,$\angle A = 90^{\circ}$.
In a $\triangle ABC$,we know the property $r_2 + r_3 = 4R \cos^2 \frac{A}{2}$.
Substituting $\angle A = 90^{\circ}$:
$r_2 + r_3 = 4R \cos^2 \left(\frac{90^{\circ}}{2}\right) = 4R \cos^2 45^{\circ}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have $\cos^2 45^{\circ} = \frac{1}{2}$.
Thus,$r_2 + r_3 = 4R \times \frac{1}{2} = 2R$.
Now,calculating the expression:
$\cos^{-1}\left(\frac{R}{r_2+r_3}\right) = \cos^{-1}\left(\frac{R}{2R}\right) = \cos^{-1}\left(\frac{1}{2}\right) = 60^{\circ}$.

(C) We use the formula for $\tan ^{-1} x + \tan ^{-1} y$. Since $x \times y = 2 \times 3 = 6 > 1$,the formula is: $\tan ^{-1} x + \tan ^{-1} y = \pi + \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
Substituting the values,we get: $\tan ^{-1} 2 + \tan ^{-1} 3 = \pi + \tan ^{-1} \left( \frac{2+3}{1-(2 \times 3)} \right)$.
$= \pi + \tan ^{-1} \left( \frac{5}{1-6} \right) = \pi + \tan ^{-1} \left( \frac{5}{-5} \right)$.
$= \pi + \tan ^{-1}(-1)$.
Since $\tan ^{-1}(-1) = -\frac{\pi}{4}$,we have: $\pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.

(A) We know that $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Given equation: $2 \operatorname{Cos}^{-1} x + \operatorname{Sin}^{-1} x = \frac{11 \pi}{6}$.
We can rewrite this as: $\operatorname{Cos}^{-1} x + (\operatorname{Cos}^{-1} x + \operatorname{Sin}^{-1} x) = \frac{11 \pi}{6}$.
Substituting the identity: $\operatorname{Cos}^{-1} x + \frac{\pi}{2} = \frac{11 \pi}{6}$.
$\operatorname{Cos}^{-1} x = \frac{11 \pi}{6} - \frac{\pi}{2} = \frac{11 \pi - 3 \pi}{6} = \frac{8 \pi}{6} = \frac{4 \pi}{3}$.
However,the range of $\operatorname{Cos}^{-1} x$ is $[0, \pi]$.
Since $\frac{4 \pi}{3} > \pi$,there is no value of $x$ that satisfies this equation.
Therefore,the number of solutions is $0$.

(A) Given: $\sec ^{-1} \left(\frac{x}{a}\right)-\sec ^{-1} \left(\frac{x}{b}\right)=\sec ^{-1} b-\sec ^{-1} a$
Using $\sec ^{-1} z = \cos ^{-1} \left(\frac{1}{z}\right)$,we get:
$\cos ^{-1} \left(\frac{a}{x}\right)-\cos ^{-1} \left(\frac{b}{x}\right)=\cos ^{-1} \left(\frac{1}{b}\right)-\cos ^{-1} \left(\frac{1}{a}\right)$
Using the formula $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} \left(uv + \sqrt{1-u^2}\sqrt{1-v^2}\right)$,we equate the arguments:
$\frac{a b}{x^2} + \sqrt{1-\frac{a^2}{x^2}}\sqrt{1-\frac{b^2}{x^2}} = \frac{1}{a b} + \sqrt{1-\frac{1}{b^2}}\sqrt{1-\frac{1}{a^2}}$
$\frac{a b}{x^2} + \frac{\sqrt{(x^2-a^2)(x^2-b^2)}}{x^2} = \frac{1}{a b} + \frac{\sqrt{(b^2-1)(a^2-1)}}{a b}$
Multiplying by $x^2 ab$:
$a^2 b^2 + ab\sqrt{(x^2-a^2)(x^2-b^2)} = x^2 + x^2\sqrt{(a^2-1)(b^2-1)}$
$ab\sqrt{(x^2-a^2)(x^2-b^2)} - x^2\sqrt{(a^2-1)(b^2-1)} = x^2 - a^2 b^2$
For the equation to hold,we set $x^2 - a^2 b^2 = 0$,which implies $x^2 = a^2 b^2$,so $x = ab$.

(A) 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 $\cot \left[\sum_{n=3}^{32} \cot ^{-1}(1+n(n+1))\right]$.
Using the identity $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$,we have $\cot^{-1}(1+n(n+1)) = \tan^{-1}\left(\frac{1}{1+n(n+1)}\right)$.
Since $\frac{1}{1+n(n+1)} = \frac{(n+1)-n}{1+(n+1)n}$,we can write this as $\tan^{-1}(n+1) - \tan^{-1}(n)$.
Thus,the sum becomes $\sum_{n=3}^{32} [\tan^{-1}(n+1) - \tan^{-1}(n)]$.
This is a telescoping sum: $(\tan^{-1} 4 - \tan^{-1} 3) + (\tan^{-1} 5 - \tan^{-1} 4) + \dots + (\tan^{-1} 33 - \tan^{-1} 32)$.
After cancellation,we are left with $\tan^{-1} 33 - \tan^{-1} 3$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$,we get $\tan^{-1}\left(\frac{33-3}{1+33 \times 3}\right) = \tan^{-1}\left(\frac{30}{100}\right) = \tan^{-1}\left(\frac{3}{10}\right)$.
Finally,we need to evaluate $\cot(\tan^{-1}(\frac{3}{10}))$.
Since $\cot(\tan^{-1}(x)) = \cot(\cot^{-1}(\frac{1}{x})) = \frac{1}{x}$,we have $\cot(\tan^{-1}(\frac{3}{10})) = \frac{10}{3}$.
Therefore,the correct option is $A$.

(D) Given $\theta = \sec^{-1}(\cosh u)$,we have $\sec \theta = \cosh u$.
Using the definition of the inverse hyperbolic cosine function,$u = \cosh^{-1}(\sec \theta) = \log_e(\sec \theta + \sqrt{\sec^2 \theta - 1})$.
Since $\sqrt{\sec^2 \theta - 1} = \tan \theta$,we get $u = \log_e(\sec \theta + \tan \theta)$.
Rewriting in terms of sine and cosine: $u = \log_e\left(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}\right) = \log_e\left(\frac{1 + \sin \theta}{\cos \theta}\right)$.
Using trigonometric identities $\sin \theta = \cos(\frac{\pi}{2} - \theta)$ and $\cos \theta = \sin(\frac{\pi}{2} - \theta)$ is not ideal here; instead,use $1 + \sin \theta = 1 + \cos(\frac{\pi}{2} - \theta) = 2\cos^2(\frac{\pi}{4} - \frac{\theta}{2})$ and $\cos \theta = \sin(\frac{\pi}{2} - \theta) = 2\sin(\frac{\pi}{4} - \frac{\theta}{2})\cos(\frac{\pi}{4} - \frac{\theta}{2})$.
Alternatively,using $1 + \sin \theta = (\cos \frac{\theta}{2} + \sin \frac{\theta}{2})^2$ and $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})$,we get $\frac{1 + \sin \theta}{\cos \theta} = \frac{\cos \frac{\theta}{2} + \sin \frac{\theta}{2}}{\cos \frac{\theta}{2} - \sin \frac{\theta}{2}} = \tan(\frac{\pi}{4} + \frac{\theta}{2})$.
Thus,$u = \log_e(\tan(\frac{\pi}{4} + \frac{\theta}{2}))$.

(A) Let $x = \sin ^{-1} \frac{12}{13} + \cos ^{-1} \frac{4}{5} + \tan ^{-1} \frac{63}{16}$.
Convert all terms to $\tan ^{-1}$ form:
$\sin ^{-1} \frac{12}{13} = \tan ^{-1} \frac{12}{5}$ (since $\sin \theta = \frac{12}{13} \implies \tan \theta = \frac{12}{5}$)
$\cos ^{-1} \frac{4}{5} = \tan ^{-1} \frac{3}{4}$ (since $\cos \theta = \frac{4}{5} \implies \tan \theta = \frac{3}{4}$)
Now,the expression becomes $\tan ^{-1} \frac{12}{5} + \tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{63}{16}$.
Using $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \frac{x+y}{1-xy}$:
$\tan ^{-1} \frac{12}{5} + \tan ^{-1} \frac{3}{4} = \tan ^{-1} \left( \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \cdot \frac{3}{4}} \right) = \tan ^{-1} \left( \frac{\frac{48+15}{20}}{1 - \frac{36}{20}} \right) = \tan ^{-1} \left( \frac{63/20}{-16/20} \right) = \tan ^{-1} \left( -\frac{63}{16} \right) = -\tan ^{-1} \frac{63}{16}$.
Substituting this back: $-\tan ^{-1} \frac{63}{16} + \tan ^{-1} \frac{63}{16} = 0$.
Wait,let us re-evaluate the sum: $\tan ^{-1} \frac{12}{5} + \tan ^{-1} \frac{3}{4} = \pi + \tan ^{-1} \left( -\frac{63}{16} \right) = \pi - \tan ^{-1} \frac{63}{16}$ (since the product $xy > 1$).
Thus,$(\pi - \tan ^{-1} \frac{63}{16}) + \tan ^{-1} \frac{63}{16} = \pi$.

(A) Given that,$\cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}$.
Since $\cos ^{-1} x = \cot ^{-1} \left(\frac{x}{\sqrt{1-x^2}}\right)$ and $\tan ^{-1} \theta = \sec ^{-1} \left(\sqrt{1+\theta^2}\right)$,we have:
$\cot \left(\cot ^{-1} \frac{x}{\sqrt{1-x^2}}\right) = \sec \left\{\sec ^{-1} \sqrt{1+\left(\frac{a}{\sqrt{b^2-a^2}}\right)^2}\right\}$
$\Rightarrow \frac{x}{\sqrt{1-x^2}} = \sqrt{1+\frac{a^2}{b^2-a^2}} = \sqrt{\frac{b^2-a^2+a^2}{b^2-a^2}} = \frac{b}{\sqrt{b^2-a^2}}$.
Squaring both sides,we get:
$\frac{x^2}{1-x^2} = \frac{b^2}{b^2-a^2}$
$x^2(b^2-a^2) = b^2(1-x^2)$
$x^2 b^2 - x^2 a^2 = b^2 - x^2 b^2$
$2 x^2 b^2 - x^2 a^2 = b^2$
$x^2(2 b^2 - a^2) = b^2$
$x^2 = \frac{b^2}{2 b^2 - a^2}$
$x = \frac{b}{\sqrt{2 b^2 - a^2}}$.

(B) Let $f(y) = y^2-4y+6 = (y-2)^2+2$. Since $(y-2)^2 \geq 0$,we have $f(y) \geq 2$ for all $y \in R$.
For statement $I$: The identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ holds if and only if $x \in [-1, 1]$. Here,$f(y) \geq 2$,so $f(y)$ is never in $[-1, 1]$. Thus,statement $I$ is false.
For statement $II$: The identity $\sec^{-1}(x) + \operatorname{cosec}^{-1}(x) = \frac{\pi}{2}$ holds if and only if $|x| \geq 1$. Since $f(y) \geq 2$,the condition $|f(y)| \geq 1$ is satisfied for all $y \in R$. Thus,statement $II$ is true.

(B) The given equation is $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$.
Both series are infinite geometric progressions with common ratios $-x/2$ and $-x^2/2$ respectively.
The sum of an infinite $G$.$P$. is $S = \frac{a}{1-r}$.
For the first series,$a=x$ and $r=-x/2$,so the sum is $\frac{x}{1-(-x/2)} = \frac{x}{1+x/2} = \frac{2x}{2+x}$.
Wait,the series given is $x - x^2/2 + x^3/4 - \dots$,which is a $G$.$P$. with $a=x$ and $r=-x/2$. The sum is $\frac{x}{1-(-x/2)} = \frac{2x}{2+x}$.
Similarly,for the second series $x^2 - x^4/2 + x^6/4 - \dots$,$a=x^2$ and $r=-x^2/2$. The sum is $\frac{x^2}{1-(-x^2/2)} = \frac{2x^2}{2+x^2}$.
Using the property $\sin ^{-1}(u) + \cos ^{-1}(v) = \pi/2$,we must have $u=v$.
Thus,$\frac{2x}{2+x} = \frac{2x^2}{2+x^2}$.
Since $x \neq 0$,we divide by $2x$: $\frac{1}{2+x} = \frac{x}{2+x^2}$.
$2+x^2 = x(2+x) \implies 2+x^2 = 2x+x^2$.
$2x = 2 \implies x=1$.

(A) Let $x = \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}$.
Note that $8-2 \sqrt{15} = (\sqrt{5}-\sqrt{3})^2$,so $\sqrt{8-2 \sqrt{15}} = \sqrt{5}-\sqrt{3}$.
Thus,$x = \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$.
Rationalizing the denominator: $x = \frac{(\sqrt{5}-\sqrt{3})^2}{5-3} = \frac{5+3-2 \sqrt{15}}{2} = \frac{8-2 \sqrt{15}}{2} = 4-\sqrt{15}$.
We know that $\operatorname{Tan}^{-1}(4-\sqrt{15}) = \operatorname{Tan}^{-1} \left( \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} \right) = \operatorname{Tan}^{-1} \left( \frac{\frac{\sqrt{5}}{\sqrt{3}}-1}{\frac{\sqrt{5}}{\sqrt{3}}+1} \right) = \operatorname{Tan}^{-1} \left( \frac{\sqrt{5}}{\sqrt{3}} \right) - \operatorname{Tan}^{-1}(1) = \operatorname{Tan}^{-1} \left( \sqrt{\frac{5}{3}} \right) - \frac{\pi}{4}$.
Alternatively,using $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} \right)$,the expression simplifies to $\operatorname{Tan}^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$.

(C) We know that $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$.
Each term can be expressed as $\tan^{-1}\left(\frac{1}{n^2+n+1}\right) = \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) = \tan^{-1}(n+1) - \tan^{-1}(n)$.
Applying this to the given terms:
$\tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(2) - \tan^{-1}(1)$
$\tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}(3) - \tan^{-1}(2)$
$\tan^{-1}\left(\frac{1}{13}\right) = \tan^{-1}(4) - \tan^{-1}(3)$
$\tan^{-1}\left(\frac{1}{21}\right) = \tan^{-1}(5) - \tan^{-1}(4)$
$\tan^{-1}\left(\frac{1}{31}\right) = \tan^{-1}(6) - \tan^{-1}(5)$
Summing these,we get a telescoping series:
$\theta = (\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + (\tan^{-1}(4) - \tan^{-1}(3)) + (\tan^{-1}(5) - \tan^{-1}(4)) + (\tan^{-1}(6) - \tan^{-1}(5))$
$\theta = \tan^{-1}(6) - \tan^{-1}(1)$
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$:
$\theta = \tan^{-1}\left(\frac{6-1}{1+6 \times 1}\right) = \tan^{-1}\left(\frac{5}{7}\right)$
Therefore,$\tan \theta = \frac{5}{7}$.

(A) We use the formula $2 \tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$.
For $x = \frac{1}{3}$,$2 \tan^{-1}\left(\frac{1}{3}\right) = \tan^{-1}\left(\frac{2(1/3)}{1-(1/3)^2}\right) = \tan^{-1}\left(\frac{2/3}{1-1/9}\right) = \tan^{-1}\left(\frac{2/3}{8/9}\right) = \tan^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) = \tan^{-1}\left(\frac{3}{4}\right)$.
Now,the expression becomes $\tan \left(\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{7}\right)\right)$.
Using the formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$,we get:
$\tan^{-1}\left(\frac{3/4 + 1/7}{1 - (3/4)(1/7)}\right) = \tan^{-1}\left(\frac{(21+4)/28}{1 - 3/28}\right) = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1)$.
Finally,$\tan(\tan^{-1}(1)) = 1$.

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

(D) Given the equation: $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$
Using the identity $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1} \left( \frac{x+2x}{1-x(2x)} \right) = \frac{\pi}{4}$
$\frac{3x}{1-2x^2} = \tan \left( \frac{\pi}{4} \right)$
$\frac{3x}{1-2x^2} = 1$
$3x = 1 - 2x^2$
$2x^2 + 3x - 1 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$
Since $x$ must be positive for $\tan ^{-1} x + \tan ^{-1} 2x$ to be $\frac{\pi}{4}$ (as $\tan ^{-1} x$ is an increasing function),we reject the negative root.
Thus,$x = \frac{\sqrt{17}-3}{4}$.

(A) Given that $\alpha = \sin^{-1} x + \cos^{-1} \left( \frac{x}{2} + \frac{\sqrt{3}}{2} \sqrt{1 - x^2} \right)$ for $0 < x < \frac{1}{2}$.
Let $x = \sin \theta$. Since $0 < x < \frac{1}{2}$,we have $0 < \theta < \frac{\pi}{6}$.
Then $\sqrt{1 - x^2} = \cos \theta$.
Substituting these into the expression for $\alpha$:
$\alpha = \sin^{-1}(\sin \theta) + \cos^{-1} \left( \frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta \right)$.
Using the identity $\cos(A - B) = \cos A \cos B + \sin A \sin B$,we have $\frac{1}{2} \sin \theta + \frac{\sqrt{3}}{2} \cos \theta = \sin \frac{\pi}{6} \sin \theta + \cos \frac{\pi}{6} \cos \theta = \cos \left( \theta - \frac{\pi}{6} \right)$.
Since $0 < \theta < \frac{\pi}{6}$,we have $-\frac{\pi}{6} < \theta - \frac{\pi}{6} < 0$,so $0 < \frac{\pi}{6} - \theta < \frac{\pi}{6}$.
Thus,$\cos^{-1} \left( \cos \left( \theta - \frac{\pi}{6} \right) \right) = \cos^{-1} \left( \cos \left( \frac{\pi}{6} - \theta \right) \right) = \frac{\pi}{6} - \theta$.
Therefore,$\alpha = \theta + \frac{\pi}{6} - \theta = \frac{\pi}{6}$.
Finally,$\tan \alpha + \cot \alpha = \tan \frac{\pi}{6} + \cot \frac{\pi}{6} = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{1 + 3}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.

(D) Given expression: $\tan^{-1}(-2) - \tan^{-1}(3)$
Using the property $\tan^{-1}(-x) = -\tan^{-1}x$,we get:
$= -\tan^{-1}(2) - \tan^{-1}(3)$
$= -(\tan^{-1}(2) + \tan^{-1}(3))$
Since $xy = 2 \times 3 = 6 > 1$,we use the formula $\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x + y}{1 - xy}\right)$:
$= -\left(\pi + \tan^{-1}\left(\frac{2 + 3}{1 - 2 \times 3}\right)\right)$
$= -\pi - \tan^{-1}\left(\frac{5}{1 - 6}\right)$
$= -\pi - \tan^{-1}(-1)$
Since $\tan^{-1}(-1) = -\frac{\pi}{4}$,we have:
$= -\pi - (-\frac{\pi}{4})$
$= -\pi + \frac{\pi}{4}$
$= -\frac{3 \pi}{4}$

(B) Given that,$\theta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$.
Using the property $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$ for $x > 0$,we have:
$\theta = \tan^{-1}(\frac{1}{7}) + \tan^{-1}(\frac{1}{8}) + \tan^{-1}(\frac{1}{18})$.
First,apply the formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})$ for the first two terms:
$\tan^{-1}(\frac{1}{7}) + \tan^{-1}(\frac{1}{8}) = \tan^{-1}(\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7 \times 8}}) = \tan^{-1}(\frac{\frac{15}{56}}{\frac{55}{56}}) = \tan^{-1}(\frac{15}{55}) = \tan^{-1}(\frac{3}{11})$.
Now,add the third term:
$\theta = \tan^{-1}(\frac{3}{11}) + \tan^{-1}(\frac{1}{18}) = \tan^{-1}(\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11 \times 18}}) = \tan^{-1}(\frac{\frac{54+11}{198}}{\frac{198-3}{198}}) = \tan^{-1}(\frac{65}{195}) = \tan^{-1}(\frac{1}{3})$.
Thus,$\tan \theta = \frac{1}{3}$,which implies $\cot \theta = 3$.

(D) We use the properties $\cot ^{-1}(-x) = \pi - \cot ^{-1} x$ and $\tan ^{-1}(-x) = -\tan ^{-1} x$ and $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$.
Given expression: $E = \tan ^{-1} 2 + \cot ^{-1}(-3) + \cot ^{-1} \frac{1}{3} + \tan ^{-1}\left(-\frac{1}{2}\right)$
Using the properties:
$E = \tan ^{-1} 2 + (\pi - \cot ^{-1} 3) + \tan ^{-1} 3 - \tan ^{-1} \frac{1}{2}$
Since $\cot ^{-1} 3 = \tan ^{-1} \frac{1}{3}$,we have:
$E = \pi + (\tan ^{-1} 2 - \tan ^{-1} \frac{1}{2}) + (\tan ^{-1} 3 - \tan ^{-1} \frac{1}{3})$
Using $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \frac{x-y}{1+xy}$:
$E = \pi + \tan ^{-1} \left( \frac{2 - 1/2}{1 + 2(1/2)} \right) + \tan ^{-1} \left( \frac{3 - 1/3}{1 + 3(1/3)} \right)$
$E = \pi + \tan ^{-1} \left( \frac{3/2}{2} \right) + \tan ^{-1} \left( \frac{8/3}{2} \right)$
$E = \pi + \tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{4}{3}$
Since $\tan ^{-1} \frac{4}{3} = \cot ^{-1} \frac{3}{4}$:
$E = \pi + (\tan ^{-1} \frac{3}{4} + \cot ^{-1} \frac{3}{4})$
Using $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$:
$E = \pi + \frac{\pi}{2} = \frac{3 \pi}{2}$

(A) Given: $\cos^{-1} \left( \frac{x}{2} \right) + \cos^{-1} \left( \frac{y}{3} \right) = \theta$
Using the identity $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$,we get:
$\cos^{-1} \left\{ \left( \frac{x}{2} \right) \left( \frac{y}{3} \right) - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}} \right\} = \theta$
$\Rightarrow \frac{xy}{6} - \sqrt{\frac{4-x^2}{4}} \sqrt{\frac{9-y^2}{9}} = \cos \theta$
$\Rightarrow \frac{xy}{6} - \frac{\sqrt{4-x^2} \sqrt{9-y^2}}{6} = \cos \theta$
$\Rightarrow xy - \sqrt{4-x^2} \sqrt{9-y^2} = 6 \cos \theta$
$\Rightarrow xy - 6 \cos \theta = \sqrt{4-x^2} \sqrt{9-y^2}$
Squaring both sides:
$(xy - 6 \cos \theta)^2 = (4-x^2)(9-y^2)$
$x^2 y^2 - 12xy \cos \theta + 36 \cos^2 \theta = 36 - 4y^2 - 9x^2 + x^2 y^2$
Canceling $x^2 y^2$ from both sides:
$-12xy \cos \theta + 36 \cos^2 \theta = 36 - 9x^2 - 4y^2$
Rearranging the terms:
$9x^2 + 4y^2 - 12xy \cos \theta = 36 - 36 \cos^2 \theta$
$9x^2 + 4y^2 - 12xy \cos \theta = 36(1 - \cos^2 \theta)$
$9x^2 + 4y^2 - 12xy \cos \theta = 36 \sin^2 \theta$

(C) Given $S_a(x) = \operatorname{Sec}^{-1}\left(\frac{x}{a}\right) + \operatorname{Sec}^{-1}(a)$ and $S_b(x) = \operatorname{Sec}^{-1}\left(\frac{x}{b}\right) + \operatorname{Sec}^{-1}(b)$.
Setting $S_a(x) = S_b(x)$,we have $\operatorname{Sec}^{-1}\left(\frac{x}{a}\right) + \operatorname{Sec}^{-1}(a) = \operatorname{Sec}^{-1}\left(\frac{x}{b}\right) + \operatorname{Sec}^{-1}(b)$.
Using the identity $\operatorname{Sec}^{-1}(y) = \cos^{-1}\left(\frac{1}{y}\right)$,we get $\cos^{-1}\left(\frac{a}{x}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{b}{x}\right) + \cos^{-1}\left(\frac{1}{b}\right)$.
If $x = ab$,then $\cos^{-1}\left(\frac{a}{ab}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{1}{b}\right) + \cos^{-1}\left(\frac{1}{a}\right)$.
This simplifies to $\cos^{-1}\left(\frac{1}{b}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{b}{ab}\right) + \cos^{-1}\left(\frac{1}{b}\right)$,which is $\cos^{-1}\left(\frac{1}{b}\right) + \cos^{-1}\left(\frac{1}{a}\right) = \cos^{-1}\left(\frac{1}{a}\right) + \cos^{-1}\left(\frac{1}{b}\right)$.
This identity holds true for $x = ab$.

(A) We know that $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$.
We can rewrite the general term as $\tan^{-1} \left( \frac{(n+1)-n}{1+n(n+1)} \right) = \tan^{-1}(n+1) - \tan^{-1} n$.
Thus,the sum is $\sum_{n=1}^{50} (\tan^{-1}(n+1) - \tan^{-1} n)$.
This is a telescoping sum: $(\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1} 51 - \tan^{-1} 50) = \tan^{-1} 51 - \tan^{-1} 1$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$,we get $\tan^{-1} \left( \frac{51-1}{1+51 \times 1} \right) = \tan^{-1} \left( \frac{50}{52} \right) = \tan^{-1} \left( \frac{25}{26} \right)$.
Finally,$\cot \left( \tan^{-1} \left( \frac{25}{26} \right) \right) = \cot \left( \cot^{-1} \left( \frac{26}{25} \right) \right) = \frac{26}{25}$.

(D) We know that $\tan ^{-1}\left[\frac{1}{1+k(k+1)}\right] = \tan ^{-1}\left[\frac{(k+1)-k}{1+k(k+1)}\right] = \tan ^{-1}(k+1) - \tan ^{-1}(k)$.
Applying this to each term in the sum:
$\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right] = \tan ^{-1}(2) - \tan ^{-1}(1)$
$\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right] = \tan ^{-1}(3) - \tan ^{-1}(2)$
...
$\tan ^{-1}\left[\frac{1}{1+n(n+1)}\right] = \tan ^{-1}(n+1) - \tan ^{-1}(n)$
Summing these terms,we get a telescoping series:
$S = (\tan ^{-1}(2) - \tan ^{-1}(1)) + (\tan ^{-1}(3) - \tan ^{-1}(2)) + \cdots + (\tan ^{-1}(n+1) - \tan ^{-1}(n))$
$S = \tan ^{-1}(n+1) - \tan ^{-1}(1)$
Using the formula $\tan ^{-1}(A) - \tan ^{-1}(B) = \tan ^{-1}\left(\frac{A-B}{1+AB}\right)$:
$S = \tan ^{-1}\left(\frac{(n+1)-1}{1+(n+1)(1)}\right) = \tan ^{-1}\left(\frac{n}{1+n+1}\right) = \tan ^{-1}\left(\frac{n}{n+2}\right)$
Given $S = \tan ^{-1}(x)$,we have $x = \frac{n}{n+2}$.

(D) We know that $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x-y}{1+xy} \right)$.
Each term in the series can be written as $\operatorname{Tan}^{-1} \left( \frac{(k+1)-k}{1+k(k+1)} \right) = \operatorname{Tan}^{-1}(k+1) - \operatorname{Tan}^{-1}(k)$.
Summing from $k=1$ to $n$:
$S = \sum_{k=1}^{n} (\operatorname{Tan}^{-1}(k+1) - \operatorname{Tan}^{-1}(k))$
$S = (\operatorname{Tan}^{-1} 2 - \operatorname{Tan}^{-1} 1) + (\operatorname{Tan}^{-1} 3 - \operatorname{Tan}^{-1} 2) + \cdots + (\operatorname{Tan}^{-1}(n+1) - \operatorname{Tan}^{-1} n)$
This is a telescoping series,so all intermediate terms cancel out:
$S = \operatorname{Tan}^{-1}(n+1) - \operatorname{Tan}^{-1}(1)$
Using the formula $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x-y}{1+xy} \right)$:
$S = \operatorname{Tan}^{-1} \left( \frac{(n+1)-1}{1+(n+1)(1)} \right) = \operatorname{Tan}^{-1} \left( \frac{n}{1+n+1} \right) = \operatorname{Tan}^{-1} \left( \frac{n}{n+2} \right)$.
Comparing this with $\operatorname{Tan}^{-1} \theta$,we get $\theta = \frac{n}{n+2}$.

(A) The general term of the series is $T_k = \operatorname{Tan}^{-1} \frac{1}{k^2+k+1}$.
We can rewrite the argument as $\frac{1}{1+k(k+1)}$.
Using the identity $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \frac{x-y}{1+xy}$,we have:
$T_k = \operatorname{Tan}^{-1} (k+1) - \operatorname{Tan}^{-1} k$.
Summing from $k=1$ to $n$:
$S_n = \sum_{k=1}^{n} (\operatorname{Tan}^{-1} (k+1) - \operatorname{Tan}^{-1} k) = \operatorname{Tan}^{-1} (n+1) - \operatorname{Tan}^{-1} (1)$.
Using the formula $\operatorname{Tan}^{-1} x - \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \frac{x-y}{1+xy}$:
$S_n = \operatorname{Tan}^{-1} \frac{(n+1)-1}{1+(n+1)(1)} = \operatorname{Tan}^{-1} \frac{n}{n+2}$.
Thus,$\theta = \frac{n}{n+2}$.

(B) Let $y = \frac{1}{2} \sin^{-1}\left(\frac{3 \sin 2\theta}{5+4 \cos 2\theta}\right)$.
Then $2y = \sin^{-1}\left(\frac{3 \sin 2\theta}{5+4 \cos 2\theta}\right)$,so $\sin 2y = \frac{3 \sin 2\theta}{5+4 \cos 2\theta}$.
Using the identity $\sin 2y = \frac{2 \tan y}{1+\tan^2 y}$ and $\sin 2\theta = \frac{2 \tan \theta}{1+\tan^2 \theta}$,$\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$,we have:
$\frac{2 \tan y}{1+\tan^2 y} = \frac{3 \left(\frac{2 \tan \theta}{1+\tan^2 \theta}\right)}{5+4 \left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)} = \frac{6 \tan \theta}{5(1+\tan^2 \theta) + 4(1-\tan^2 \theta)} = \frac{6 \tan \theta}{9+\tan^2 \theta}$.
Let $t = \tan \theta$ and $x = \tan y$. Then $\frac{2x}{1+x^2} = \frac{6t}{9+t^2}$.
Cross-multiplying: $2x(9+t^2) = 6t(1+x^2) \implies 18x + 2xt^2 = 6t + 6tx^2$.
Rearranging: $6tx^2 - (18+2t^2)x + 6t = 0 \implies 3tx^2 - (9+t^2)x + 3t = 0$.
Factoring: $(3x-t)(tx-3) = 0$.
Thus,$x = \frac{t}{3} = \frac{1}{3} \tan \theta$.

(C) We can rewrite each term using the formula $\tan^{-1} \frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1} b$.
$y = \tan^{-1} \frac{2x-x}{1+(2x)(x)} + \tan^{-1} \frac{3x-2x}{1+(3x)(2x)} + \tan^{-1} \frac{4x-3x}{1+(4x)(3x)}$
$y = (\tan^{-1} 2x - \tan^{-1} x) + (\tan^{-1} 3x - \tan^{-1} 2x) + (\tan^{-1} 4x - \tan^{-1} 3x)$
$y = \tan^{-1} 4x - \tan^{-1} x$
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{4}{1+(4x)^2} - \frac{1}{1+x^2} = \frac{4}{1+16x^2} - \frac{1}{1+x^2}$
Substitute $x = \frac{1}{2}$:
$\left(\frac{dy}{dx}\right)_{x=\frac{1}{2}} = \frac{4}{1+16(\frac{1}{4})} - \frac{1}{1+(\frac{1}{4})} = \frac{4}{1+4} - \frac{1}{\frac{5}{4}} = \frac{4}{5} - \frac{4}{5} = 0$