Properties of ITF Questions in English

Let $\sin ^{-1} \frac{5}{13} = x$.
Then,$\sin x = \frac{5}{13}$,which implies $\cos x = \sqrt{1 - (\frac{5}{13})^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
Therefore,$\tan x = \frac{\sin x}{\cos x} = \frac{5/13}{12/13} = \frac{5}{12}$,so $x = \tan ^{-1} \frac{5}{12}$.
Thus,$\sin ^{-1} \frac{5}{13} = \tan ^{-1} \frac{5}{12}$ $\ldots (1)$.
Let $\cos ^{-1} \frac{3}{5} = y$.
Then,$\cos y = \frac{3}{5}$,which implies $\sin y = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Therefore,$\tan y = \frac{\sin y}{\cos y} = \frac{4/5}{3/5} = \frac{4}{3}$,so $y = \tan ^{-1} \frac{4}{3}$.
Thus,$\cos ^{-1} \frac{3}{5} = \tan ^{-1} \frac{4}{3}$ $\ldots (2)$.
Now,consider the $R$.$H$.$S$.:
$\sin ^{-1} \frac{5}{13} + \cos ^{-1} \frac{3}{5} = \tan ^{-1} \frac{5}{12} + \tan ^{-1} \frac{4}{3}$.
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} (\frac{a+b}{1-ab})$:
$= \tan ^{-1} (\frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \times \frac{4}{3}}) = \tan ^{-1} (\frac{\frac{15+48}{36}}{1 - \frac{20}{36}}) = \tan ^{-1} (\frac{63/36}{16/36}) = \tan ^{-1} \frac{63}{16}$.
$= L.H.S.$
Hence,the identity is proved.

(N/A) $L$.$H$.$S$ = $\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}$
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$,we group the terms:
$= \left( \tan ^{-1} \frac{1}{5} + \tan ^{-1} \frac{1}{7} \right) + \left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{8} \right)$
$= \tan ^{-1} \left( \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \times \frac{1}{7}} \right) + \tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \times \frac{1}{8}} \right)$
$= \tan ^{-1} \left( \frac{\frac{12}{35}}{\frac{34}{35}} \right) + \tan ^{-1} \left( \frac{\frac{11}{24}}{\frac{23}{24}} \right)$
$= \tan ^{-1} \left( \frac{12}{34} \right) + \tan ^{-1} \left( \frac{11}{23} \right) = \tan ^{-1} \left( \frac{6}{17} \right) + \tan ^{-1} \left( \frac{11}{23} \right)$
$= \tan ^{-1} \left( \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \times \frac{11}{23}} \right) = \tan ^{-1} \left( \frac{\frac{138 + 187}{391}}{\frac{391 - 66}{391}} \right)$
$= \tan ^{-1} \left( \frac{325}{325} \right) = \tan ^{-1} (1) = \frac{\pi}{4} = R.H.S$

Let $x = \tan^2 \theta$.
Then $\sqrt{x} = \tan \theta$,which implies $\theta = \tan^{-1} \sqrt{x}$.
Consider the expression $\frac{1-x}{1+x}$. Substituting $x = \tan^2 \theta$,we get:
$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos(2\theta)$.
Now,consider the Right Hand Side $(RHS)$:
$RHS = \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right)$
$= \frac{1}{2} \cos^{-1} (\cos 2\theta)$
$= \frac{1}{2} \times 2\theta = \theta$.
Since $\theta = \tan^{-1} \sqrt{x}$,we have $RHS = \tan^{-1} \sqrt{x} = LHS$.
Hence,the identity is proved.

Let $y = \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$.
Rationalizing the denominator,we multiply the numerator and denominator by $(\sqrt{1+\sin x}+\sqrt{1-\sin x})$:
$y = \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$
$y = \frac{(1+\sin x) + (1-\sin x) + 2\sqrt{(1+\sin x)(1-\sin x)}}{(1+\sin x) - (1-\sin x)}$
$y = \frac{2 + 2\sqrt{1-\sin^2 x}}{2\sin x} = \frac{2 + 2\cos x}{2\sin x} = \frac{1+\cos x}{\sin x}$
Using trigonometric identities $1+\cos x = 2\cos^2 \frac{x}{2}$ and $\sin x = 2\sin \frac{x}{2}\cos \frac{x}{2}$:
$y = \frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \cot \frac{x}{2}$
Therefore,$\cot^{-1}(y) = \cot^{-1}(\cot \frac{x}{2}) = \frac{x}{2}$,which is the $R.H.S.$

(A) Let $x = \cos 2\theta$,then $\theta = \frac{1}{2} \cos ^{-1} x$.
$L.H.S. = \tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$
Substitute $x = \cos 2\theta$:
$= \tan ^{-1}\left(\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right)$
Using the identities $1+\cos 2\theta = 2\cos^2 \theta$ and $1-\cos 2\theta = 2\sin^2 \theta$:
$= \tan ^{-1}\left(\frac{\sqrt{2\cos^2 \theta}-\sqrt{2\sin^2 \theta}}{\sqrt{2\cos^2 \theta}+\sqrt{2\sin^2 \theta}}\right)$
$= \tan ^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right)$
Divide numerator and denominator by $\sqrt{2}\cos \theta$:
$= \tan ^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right)$
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$= \tan ^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right)$
$= \frac{\pi}{4} - \theta$
Substituting $\theta = \frac{1}{2} \cos ^{-1} x$:
$= \frac{\pi}{4} - \frac{1}{2} \cos ^{-1} x = R.H.S.$

(A) Consider the $L.H.S. = \frac{9 \pi}{8} - \frac{9}{4} \sin^{-1} \frac{1}{3}$.
Factor out $\frac{9}{4}$ from the expression:
$L.H.S. = \frac{9}{4} \left( \frac{\pi}{2} - \sin^{-1} \frac{1}{3} \right)$.
Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,we have $\frac{\pi}{2} - \sin^{-1} \frac{1}{3} = \cos^{-1} \frac{1}{3}$.
So,$L.H.S. = \frac{9}{4} \cos^{-1} \frac{1}{3}$.
Let $\cos^{-1} \frac{1}{3} = \theta$. Then $\cos \theta = \frac{1}{3}$.
Since $\sin^2 \theta = 1 - \cos^2 \theta$,we have $\sin^2 \theta = 1 - (\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
Thus,$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1} \frac{2 \sqrt{2}}{3}$.
Substituting this back into the expression,we get $L.H.S. = \frac{9}{4} \sin^{-1} \frac{2 \sqrt{2}}{3} = R.H.S.$

(B) Given the equation: $\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x$
Using the identity $\tan ^{-1} a - \tan ^{-1} b = \tan ^{-1} \frac{a-b}{1+ab}$,we can write $\tan ^{-1} 1 - \tan ^{-1} x = \frac{1}{2} \tan ^{-1} x$.
Since $\tan ^{-1} 1 = \frac{\pi}{4}$,the equation becomes $\frac{\pi}{4} - \tan ^{-1} x = \frac{1}{2} \tan ^{-1} x$.
Rearranging the terms,we get $\frac{\pi}{4} = \tan ^{-1} x + \frac{1}{2} \tan ^{-1} x = \frac{3}{2} \tan ^{-1} x$.
Multiplying both sides by $\frac{2}{3}$,we get $\tan ^{-1} x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.
Taking the tangent of both sides,$x = \tan \frac{\pi}{6}$.
Therefore,$x = \frac{1}{\sqrt{3}}$.

(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 $B = \frac{x-y}{x+y}$.
$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) = \tan ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right]$
$= \tan ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right]$
$= \tan ^{-1}\left(\frac{x^{2}+xy-xy+y^{2}}{xy+y^{2}+x^{2}-xy}\right)$
$= \tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right) = \tan ^{-1}(1) = \frac{\pi}{4}$.

(A) Given $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$.
Let $x = \tan \theta$,then $\theta = \tan^{-1} x$.
Substituting $x = \tan \theta$ in the expression,we get:
$y = \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right)$
Using the trigonometric identity $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$,we have:
$y = \sin^{-1}(\sin 2\theta)$
Since $|x| < 1$,$2\theta$ lies in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$,so $y = 2\theta$.
Substituting back $\theta = \tan^{-1} x$:
$y = 2 \tan^{-1} x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1} x)$
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$.

(A) Given $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Let $x = \tan \theta$,where $\theta = \tan^{-1} x$. Since $0 < x < 1$,we have $0 < \theta < \frac{\pi}{4}$.
Substituting $x = \tan \theta$ into the expression:
$y = \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)$
Using the trigonometric identity $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$:
$y = \cos^{-1}(\cos 2\theta)$
Since $0 < \theta < \frac{\pi}{4}$,it follows that $0 < 2\theta < \frac{\pi}{2}$.
Thus,$y = 2\theta = 2 \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1} x)$
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$.

(A) Given $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$.
Let $x = \tan \theta$,where $\theta = \tan^{-1} x$. Since $-1 < x < 1$,we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.
Then $\frac{2x}{1+x^2} = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin(2\theta)$.
So,$y = \cos^{-1}(\sin(2\theta)) = \cos^{-1}(\cos(\frac{\pi}{2} - 2\theta))$.
Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,we have $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$,which implies $0 < \frac{\pi}{2} - 2\theta < \pi$.
Thus,$y = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2 \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} = \frac{-2}{1+x^2}$.

(A) Given $y = \sin^{-1}(2x\sqrt{1-x^2})$.
Let $x = \sin \theta$,then $\theta = \sin^{-1} x$.
Since $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$,we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.
Then $y = \sin^{-1}(2 \sin \theta \sqrt{1-\sin^2 \theta}) = \sin^{-1}(2 \sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta)$.
Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$,we have $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$.
Thus,$y = 2\theta = 2\sin^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{2}{\sqrt{1-x^2}}$.

(B) Given $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$.
Using the identity $\sec^{-1}(u) = \cos^{-1}(1/u)$,we have:
$y = \cos^{-1}(2x^2 - 1)$.
Let $x = \cos \theta$,then $2x^2 - 1 = 2\cos^2 \theta - 1 = \cos(2\theta)$.
Since $0 < x < \frac{1}{\sqrt{2}}$,we have $\frac{\pi}{4} < \theta < \frac{\pi}{2}$,so $\frac{\pi}{2} < 2\theta < \pi$.
Thus,$y = \cos^{-1}(\cos 2\theta) = 2\theta = 2\cos^{-1}x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \left(\frac{-1}{\sqrt{1-x^2}}\right) = \frac{-2}{\sqrt{1-x^2}}$.

(A) Given $y = \sin^{-1}x + \sin^{-1}\sqrt{1-x^2}$.
Let $x = \sin\theta$,then $\theta = \sin^{-1}x$. Since $-1 \le x \le 1$,we have $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.
Then $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = |\cos\theta|$.
Since $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$,$\cos\theta \ge 0$,so $|\cos\theta| = \cos\theta$.
Thus,$y = \sin^{-1}(\sin\theta) + \sin^{-1}(\cos\theta) = \theta + \sin^{-1}(\sin(\frac{\pi}{2} - \theta))$.
For $x \in [0, 1]$,$0 \le \theta \le \frac{\pi}{2}$,so $y = \theta + (\frac{\pi}{2} - \theta) = \frac{\pi}{2}$.
For $x \in [-1, 0]$,$-\frac{\pi}{2} \le \theta < 0$,then $\cos\theta = \sin(\frac{\pi}{2} + \theta)$.
$y = \theta + \sin^{-1}(\sin(\frac{\pi}{2} + \theta))$. Since $\frac{\pi}{2} + \theta \in [0, \pi]$,we use the identity $\sin^{-1}(\sin A) = \pi - A$ for $A \in [\frac{\pi}{2}, \pi]$.
$y = \theta + \pi - (\frac{\pi}{2} + \theta) = \frac{\pi}{2}$.
In both cases,$y = \frac{\pi}{2}$.
Therefore,$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2}) = 0$.

(C) We have $S_{k} = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^{r}}{2 \cdot 2^{2r} + 3 \cdot 3^{2r}}\right)$.
Dividing the numerator and denominator by $3^{2r}$,we get:
$S_{k} = \sum_{r=1}^{k} \tan^{-1}\left(\frac{\frac{2^r}{3^r} \cdot 3^r}{2 \cdot \frac{2^{2r}}{3^{2r}} + 3}\right) = \sum_{r=1}^{k} \tan^{-1}\left(\frac{3^r \cdot (2/3)^r}{2 \cdot (2/3)^{2r} + 3}\right)$.
Actually,rewrite the term inside $\tan^{-1}$ as:
$\frac{6^r}{2 \cdot 4^r + 3 \cdot 9^r} = \frac{6^r / 9^r}{2 \cdot (4/9)^r + 3} = \frac{(2/3)^r}{2 \cdot (2/3)^{2r} + 3}$.
This can be written as $\frac{(2/3)^r / 3}{1 + \frac{2}{3} (2/3)^{2r}} = \frac{(2/3)^r - (2/3)^{r+1}}{1 + (2/3)^r \cdot (2/3)^{r+1}}$.
Thus,the sum is a telescoping series:
$S_{k} = \sum_{r=1}^{k} \left( \tan^{-1}((2/3)^r) - \tan^{-1}((2/3)^{r+1}) \right)$.
$S_{k} = \left( \tan^{-1}(2/3) - \tan^{-1}((2/3)^2) \right) + \left( \tan^{-1}((2/3)^2) - \tan^{-1}((2/3)^3) \right) + \dots + \left( \tan^{-1}((2/3)^k) - \tan^{-1}((2/3)^{k+1}) \right)$.
$S_{k} = \tan^{-1}(2/3) - \tan^{-1}((2/3)^{k+1})$.
Taking the limit as $k \rightarrow \infty$,$(2/3)^{k+1} \rightarrow 0$.
$S_{\infty} = \tan^{-1}(2/3) - \tan^{-1}(0) = \tan^{-1}(2/3) = \cot^{-1}(3/2)$.

(A) The given series is $\sum_{n=1}^{100} \cot^{-1}(2n^2)$.
Using the identity $\cot^{-1}(x) = \tan^{-1}(\frac{1}{x})$,we have $\sum_{n=1}^{100} \tan^{-1}(\frac{1}{2n^2})$.
Multiplying numerator and denominator by $2$,we get $\sum_{n=1}^{100} \tan^{-1}(\frac{2}{4n^2})$.
This can be written as $\sum_{n=1}^{100} \tan^{-1}(\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)})$.
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}(\frac{x-y}{1+xy})$,the sum becomes $\sum_{n=1}^{100} (\tan^{-1}(2n+1) - \tan^{-1}(2n-1))$.
This is a telescoping series: $(\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1} 201 - \tan^{-1} 199)$.
After cancellation,we are left with $\tan^{-1} 201 - \tan^{-1} 1$.
Using the formula again,$\tan^{-1}(\frac{201-1}{1+201 \times 1}) = \tan^{-1}(\frac{200}{202}) = \tan^{-1}(\frac{100}{101})$.
Since $\cot^{-1}(\alpha) = \tan^{-1}(\frac{101}{100})$,we have $\alpha = \frac{101}{100} = 1.01$.

(B) Given equation: $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$.
Using the property $\cot ^{-1}(u) = \tan ^{-1}\left(\frac{1}{u}\right)$ for $u > 0$,we have $\cot ^{-1}\left(\frac{1}{x-1}\right) = \tan ^{-1}(x-1)$ if $x-1 > 0$,i.e.,$x > 1$.
If $x > 1$,then $\tan ^{-1}(x+1) + \tan ^{-1}(x-1) = \tan ^{-1}\left(\frac{8}{31}\right)$.
Applying $\tan(A+B)$ formula: $\frac{(x+1)+(x-1)}{1-(x+1)(x-1)} = \frac{8}{31} \Rightarrow \frac{2x}{1-(x^2-1)} = \frac{8}{31} \Rightarrow \frac{2x}{2-x^2} = \frac{8}{31}$.
$62x = 16 - 8x^2 \Rightarrow 8x^2 + 62x - 16 = 0 \Rightarrow 4x^2 + 31x - 8 = 0$.
Solving for $x$: $(4x-1)(x+8) = 0$,so $x = \frac{1}{4}$ or $x = -8$. Both values are $\leq 1$,so the assumption $x > 1$ is invalid.
If $x < 1$,then $\cot ^{-1}\left(\frac{1}{x-1}\right) = \pi + \tan ^{-1}(x-1)$.
Then $\tan ^{-1}(x+1) + \pi + \tan ^{-1}(x-1) = \tan ^{-1}\left(\frac{8}{31}\right)$.
$\tan ^{-1}\left(\frac{2x}{2-x^2}\right) = \tan ^{-1}\left(\frac{8}{31}\right) - \pi$.
Taking $\tan$ on both sides: $\frac{2x}{2-x^2} = \tan(\tan ^{-1}(\frac{8}{31}) - \pi) = \frac{8}{31}$.
This leads to $4x^2 + 31x - 8 = 0$,giving $x = \frac{1}{4}$ or $x = -8$. Since $x < 1$,both are candidates. Checking $x = \frac{1}{4}$: $\tan^{-1}(1.25) + \cot^{-1}(1/(-0.75)) = \tan^{-1}(1.25) + \cot^{-1}(-1.33) \approx 0.89 + 2.21 > \pi/2$. Thus $x = 1/4$ is rejected.
For $x = -8$: $\tan^{-1}(-7) + \cot^{-1}(-1/9) = -1.42 + 1.68 = 0.26 = \tan^{-1}(8/31)$.
Thus,the only solution is $x = -8$.

(C) Given $\frac{\sin ^{-1} x}{a} = \frac{\cos ^{-1} x}{b} = \frac{\tan ^{-1} y}{c} = k$ (where $k$ is a constant).
Then $\sin ^{-1} x = ak$,$\cos ^{-1} x = bk$,and $\tan ^{-1} y = ck$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.
Substituting the values,we get $ak + bk = \frac{\pi}{2}$,which implies $k(a + b) = \frac{\pi}{2}$,or $a + b = \frac{\pi}{2k}$.
Now,we need to evaluate $\cos \left(\frac{\pi c}{a + b}\right)$.
Substituting $a + b = \frac{\pi}{2k}$,we get $\cos \left(\frac{\pi c}{\frac{\pi}{2k}}\right) = \cos (2ck)$.
Since $\tan ^{-1} y = ck$,we have $\cos (2ck) = \cos (2 \tan ^{-1} y)$.
Using the formula $\cos (2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$,where $\theta = \tan ^{-1} y$,we get $\cos (2 \tan ^{-1} y) = \frac{1 - y^2}{1 + y^2}$.

(A) Let $\frac{1}{4} \sin ^{-1} \frac{\sqrt{63}}{8} = \theta$.
Then,$\sin 4\theta = \frac{\sqrt{63}}{8}$.
Since $\cos^2 4\theta = 1 - \sin^2 4\theta = 1 - \frac{63}{64} = \frac{1}{64}$,we have $\cos 4\theta = \frac{1}{8}$.
Using the formula $\cos 4\theta = 2\cos^2 2\theta - 1$,we get $2\cos^2 2\theta - 1 = \frac{1}{8}$,which implies $2\cos^2 2\theta = \frac{9}{8}$,so $\cos^2 2\theta = \frac{9}{16}$.
Thus,$\cos 2\theta = \frac{3}{4}$.
Using the formula $\cos 2\theta = 2\cos^2 \theta - 1$,we get $2\cos^2 \theta - 1 = \frac{3}{4}$,which implies $2\cos^2 \theta = \frac{7}{4}$,so $\cos^2 \theta = \frac{7}{8}$.
Then $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{7}{8} = \frac{1}{8}$.
Therefore,$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1/8}{7/8} = \frac{1}{7}$.
Hence,$\tan \theta = \frac{1}{\sqrt{7}}$.

(B) Let $x = 2 \cot^{-1}(5) + \cos^{-1}(\frac{4}{5})$.
We know that $\cot^{-1}(5) = \tan^{-1}(\frac{1}{5})$.
So,$2 \cot^{-1}(5) = 2 \tan^{-1}(\frac{1}{5}) = \tan^{-1}(\frac{2(1/5)}{1-(1/5)^2}) = \tan^{-1}(\frac{2/5}{24/25}) = \tan^{-1}(\frac{2}{5} \times \frac{25}{24}) = \tan^{-1}(\frac{5}{12})$.
Also,$\cos^{-1}(\frac{4}{5}) = \tan^{-1}(\frac{3}{4})$ because if $\cos \theta = \frac{4}{5}$,then $\tan \theta = \frac{3}{4}$.
Thus,the expression becomes $\operatorname{cosec}[\tan^{-1}(\frac{5}{12}) + \tan^{-1}(\frac{3}{4})]$.
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1}(\frac{A+B}{1-AB})$:
$\tan^{-1}(\frac{5/12 + 3/4}{1 - (5/12)(3/4)}) = \tan^{-1}(\frac{(5+9)/12}{1 - 15/48}) = \tan^{-1}(\frac{14/12}{33/48}) = \tan^{-1}(\frac{14}{12} \times \frac{48}{33}) = \tan^{-1}(\frac{56}{33})$.
Finally,$\operatorname{cosec}(\tan^{-1}(\frac{56}{33}))$. Let $\theta = \tan^{-1}(\frac{56}{33})$,then $\tan \theta = \frac{56}{33}$.
Since $\operatorname{cosec} \theta = \sqrt{1 + \cot^2 \theta} = \sqrt{1 + (33/56)^2} = \sqrt{\frac{56^2 + 33^2}{56^2}} = \sqrt{\frac{3136 + 1089}{3136}} = \sqrt{\frac{4225}{3136}} = \frac{65}{56}$.

(B) Given $a = (\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}$.
Using the identity $A^2 - B^2 = (A+B)(A-B)$,we get:
$a = (\sin ^{-1} x + \cos ^{-1} x)(\sin ^{-1} x - \cos ^{-1} x)$.
Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have:
$a = \frac{\pi}{2}(\sin ^{-1} x - \cos ^{-1} x)$.
Substitute $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$:
$a = \frac{\pi}{2}(\frac{\pi}{2} - 2 \cos ^{-1} x)$.
$\frac{2a}{\pi} = \frac{\pi}{2} - 2 \cos ^{-1} x$.
$2 \cos ^{-1} x = \frac{\pi}{2} - \frac{2a}{\pi}$.
Taking cosine on both sides:
$\cos(2 \cos ^{-1} x) = \cos(\frac{\pi}{2} - \frac{2a}{\pi})$.
Using the identity $\cos(2 \theta) = 2 \cos^2 \theta - 1$ and $\cos(\frac{\pi}{2} - \theta) = \sin \theta$:
$2 x^2 - 1 = \sin(\frac{2a}{\pi})$.

(D) Let $g(x) = \sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$.
For $x \in [0, \frac{\pi}{2}]$,the argument $(x + \frac{\pi}{4}) \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.
Thus,$g(x) \in [1, \sqrt{2}]$.
Since $f(x) = \tan^{-1}(g(x))$ is an increasing function,the range of $f(x)$ is $[\tan^{-1}(1), \tan^{-1}(\sqrt{2})] = [\frac{\pi}{4}, \tan^{-1}(\sqrt{2})]$.
Therefore,$m = \frac{\pi}{4}$ and $M = \tan^{-1}(\sqrt{2})$.
We need to find $\tan(M - m) = \tan(\tan^{-1}(\sqrt{2}) - \frac{\pi}{4})$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,we get:
$\tan(M - m) = \frac{\sqrt{2} - 1}{1 + \sqrt{2}(1)} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$.
Rationalizing the denominator:
$\frac{\sqrt{2} - 1}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{2 - 2\sqrt{2} + 1}{2 - 1} = 3 - 2\sqrt{2}$.

(D) Let $\alpha = \tan^{-1}\left(\frac{3}{5}\right)$ and $\beta = \sin^{-1}\left(\frac{5}{13}\right)$.
We need to find $\tan(2\alpha + \beta)$.
First,convert $\beta$ to $\tan^{-1}$ form: Since $\sin \beta = \frac{5}{13}$,then $\tan \beta = \frac{5}{\sqrt{13^2 - 5^2}} = \frac{5}{12}$. So,$\beta = \tan^{-1}\left(\frac{5}{12}\right)$.
Now,$2\alpha = 2\tan^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{3}{5}}{1 - (\frac{3}{5})^2}\right) = \tan^{-1}\left(\frac{6/5}{1 - 9/25}\right) = \tan^{-1}\left(\frac{6/5}{16/25}\right) = \tan^{-1}\left(\frac{6}{5} \cdot \frac{25}{16}\right) = \tan^{-1}\left(\frac{15}{8}\right)$.
Now we need $\tan(2\alpha + \beta) = \tan\left(\tan^{-1}\left(\frac{15}{8}\right) + \tan^{-1}\left(\frac{5}{12}\right)\right)$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$:
$= \tan\left(\tan^{-1}\left(\frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}}\right)\right) = \frac{\frac{45+10}{24}}{1 - \frac{75}{96}} = \frac{55/24}{21/96} = \frac{55}{24} \cdot \frac{96}{21} = \frac{55 \cdot 4}{21} = \frac{220}{21}$.

(B) Given the operation $x * y = x^{2} + y^{3}$.
First,evaluate $(x * 1) * 1 = x * (1 * 1)$:
$(x * 1) = x^{2} + 1^{3} = x^{2} + 1$.
So,$(x * 1) * 1 = (x^{2} + 1) * 1 = (x^{2} + 1)^{2} + 1^{3} = (x^{2} + 1)^{2} + 1$.
Now,evaluate $x * (1 * 1)$:
$(1 * 1) = 1^{2} + 1^{3} = 2$.
So,$x * (1 * 1) = x * 2 = x^{2} + 2^{3} = x^{2} + 8$.
Equating both sides:
$(x^{2} + 1)^{2} + 1 = x^{2} + 8$
$x^{4} + 2x^{2} + 1 + 1 = x^{2} + 8$
$x^{4} + x^{2} - 6 = 0$.
Let $t = x^{2}$,then $t^{2} + t - 6 = 0$,which factors as $(t + 3)(t - 2) = 0$.
Since $x^{2} = t$ must be non-negative,$x^{2} = 2$.
Now substitute $x^{2} = 2$ into the expression $2 \sin^{-1}\left(\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}\right)$:
$x^{4} = (x^{2})^{2} = 2^{2} = 4$.
Expression $= 2 \sin^{-1}\left(\frac{4 + 2 - 2}{4 + 2 + 2}\right) = 2 \sin^{-1}\left(\frac{4}{8}\right) = 2 \sin^{-1}\left(\frac{1}{2}\right)$.
Since $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$,the value is $2 \times \frac{\pi}{6} = \frac{\pi}{3}$.

(B) Given $y = \tan^{-1}(\sec x^3 - \tan x^3)$.
$y = \tan^{-1}\left(\frac{1 - \sin x^3}{\cos x^3}\right) = \tan^{-1}\left(\frac{1 - \cos(\frac{\pi}{2} - x^3)}{\sin(\frac{\pi}{2} - x^3)}\right)$.
Using the half-angle formulas,$1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$ and $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$,we get:
$y = \tan^{-1}\left(\tan(\frac{\pi}{4} - \frac{x^3}{2})\right)$.
Since $\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$,then $-\frac{3\pi}{4} < -\frac{x^3}{2} < -\frac{\pi}{4}$.
Adding $\frac{\pi}{4}$,we get $-\frac{\pi}{2} < \frac{\pi}{4} - \frac{x^3}{2} < 0$.
Thus,$y = \frac{\pi}{4} - \frac{x^3}{2}$.
Now,differentiate with respect to $x$:
$y' = -\frac{1}{2} \cdot 3x^2 = -\frac{3}{2}x^2$.
$y'' = -\frac{3}{2} \cdot 2x = -3x$.
From $y = \frac{\pi}{4} - \frac{x^3}{2}$,we have $x^3 = \frac{\pi}{2} - 2y$.
Also,$x^2 = -\frac{2}{3} y'$.
Substituting $x^2$ into $y'' = -3x$,we get $y'' = -3 \sqrt[3]{-\frac{2}{3} y'}$.
Alternatively,using $x^2 = -\frac{2}{3} y'$,we have $x^2 y'' = x^2 (-3x) = -3x^3 = -3(\frac{\pi}{2} - 2y) = -\frac{3\pi}{2} + 6y$.
Therefore,$x^2 y'' - 6y + \frac{3\pi}{2} = 0$.

(A) Let $f(x) = (\tan^{-1} x)^3 + (\cot^{-1} x)^3$.
We know that $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
Let $u = \tan^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$.
Since $x \in \mathbb{R}$,$u \in (-\frac{\pi}{2}, \frac{\pi}{2})$.
Now,$f(x) = u^3 + (\frac{\pi}{2} - u)^3 = u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8}$.
Completing the square: $f(x) = \frac{3\pi}{2}(u^2 - \frac{\pi}{2}u) + \frac{\pi^3}{8} = \frac{3\pi}{2}(u - \frac{\pi}{4})^2 - \frac{3\pi}{2}(\frac{\pi^2}{16}) + \frac{\pi^3}{8} = \frac{3\pi}{2}(u - \frac{\pi}{4})^2 + \frac{\pi^3}{32}$.
Since $u \in (-\frac{\pi}{2}, \frac{\pi}{2})$,the range of $(u - \frac{\pi}{4})^2$ is $[0, (-\frac{\pi}{2} - \frac{\pi}{4})^2) = [0, \frac{9\pi^2}{16})$.
Thus,the range of $f(x)$ is $[\frac{\pi^3}{32}, \frac{3\pi}{2}(\frac{9\pi^2}{16}) + \frac{\pi^3}{32}) = [\frac{\pi^3}{32}, \frac{27\pi^3}{32} + \frac{\pi^3}{32}) = [\frac{\pi^3}{32}, \frac{28\pi^3}{32}) = [\frac{\pi^3}{32}, \frac{7\pi^3}{8})$.
Given $f(x) = k \pi^3$,we have $k \in [\frac{1}{32}, \frac{7}{8})$.

(A) We know that $\tan ^{-1} \left(\frac{x-y}{1+xy}\right) = \tan ^{-1} x - \tan ^{-1} y$.
Given the term $\tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)$,we can rewrite it as:
$\tan ^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) = \tan ^{-1}(n+1) - \tan ^{-1} n$.
Now,consider the summation:
$\sum\limits_{n=1}^{50} \left(\tan ^{-1}(n+1) - \tan ^{-1} n\right) = (\tan ^{-1} 2 - \tan ^{-1} 1) + (\tan ^{-1} 3 - \tan ^{-1} 2) + \dots + (\tan ^{-1} 51 - \tan ^{-1} 50)$.
This is a telescoping series,so it simplifies to:
$\tan ^{-1} 51 - \tan ^{-1} 1$.
Using the formula $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \left(\frac{x-y}{1+xy}\right)$,we get:
$\tan ^{-1} 51 - \tan ^{-1} 1 = \tan ^{-1} \left(\frac{51-1}{1+51 \times 1}\right) = \tan ^{-1} \left(\frac{50}{52}\right) = \tan ^{-1} \left(\frac{25}{26}\right)$.
Finally,we need to find the value of $\cot \left(\tan ^{-1} \left(\frac{25}{26}\right)\right)$.
Since $\cot(\tan ^{-1} x) = \frac{1}{x}$,we have:
$\cot \left(\tan ^{-1} \left(\frac{25}{26}\right)\right) = \frac{26}{25}$.

(C) We evaluate each term individually using the principal value branches of the inverse trigonometric functions:
$1$. For $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$:
Since $\frac{2 \pi}{3}$ is not in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we write $\sin \frac{2 \pi}{3} = \sin(\pi - \frac{\pi}{3}) = \sin \frac{\pi}{3}$.
Thus,$\sin ^{-1}\left(\sin \frac{\pi}{3}\right) = \frac{\pi}{3}$.
$2$. For $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$:
Since $\frac{7 \pi}{6}$ is not in the principal range $[0, \pi]$,we write $\cos \frac{7 \pi}{6} = \cos(2\pi - \frac{5\pi}{6}) = \cos \frac{5 \pi}{6}$.
Thus,$\cos ^{-1}\left(\cos \frac{5 \pi}{6}\right) = \frac{5 \pi}{6}$.
$3$. For $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$:
Since $\frac{3 \pi}{4}$ is not in the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$,we write $\tan \frac{3 \pi}{4} = \tan(\pi - \frac{\pi}{4}) = -\tan \frac{\pi}{4} = \tan(-\frac{\pi}{4})$.
Thus,$\tan ^{-1}\left(\tan(-\frac{\pi}{4})\right) = -\frac{\pi}{4}$.
Summing these values:
$\frac{\pi}{3} + \frac{5 \pi}{6} - \frac{\pi}{4} = \frac{4\pi + 10\pi - 3\pi}{12} = \frac{11 \pi}{12}$.

(B) Given $\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta} = k$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.
Substituting the values,$k \alpha + k \beta = \frac{\pi}{2}$,which implies $k(\alpha + \beta) = \frac{\pi}{2}$,so $k = \frac{\pi}{2(\alpha + \beta)}$.
From $\sin ^{-1} x = k \alpha$,we have $\alpha = \frac{\sin ^{-1} x}{k} = \frac{2(\alpha + \beta) \sin ^{-1} x}{\pi}$.
Therefore,$\frac{\alpha}{\alpha + \beta} = \frac{2 \sin ^{-1} x}{\pi}$.
We need to find $\sin \left(\frac{2 \pi \alpha}{\alpha + \beta}\right)$.
Substituting the ratio,$\sin \left(2 \pi \cdot \frac{2 \sin ^{-1} x}{\pi}\right) = \sin (4 \sin ^{-1} x)$.
Using the formula $\sin(4\theta) = 2 \sin(2\theta) \cos(2\theta) = 2(2 \sin \theta \cos \theta)(1 - 2 \sin^2 \theta) = 4 \sin \theta \cos \theta (1 - 2 \sin^2 \theta)$.
Let $\theta = \sin ^{-1} x$,then $\sin \theta = x$ and $\cos \theta = \sqrt{1 - x^2}$.
Substituting these,we get $4 x \sqrt{1 - x^2} (1 - 2 x^2)$.

(B) Let the expression be $E = \tan \left(2 \tan ^{-1} \frac{1}{5} + 2 \tan ^{-1} \frac{1}{8} + \sec ^{-1} \frac{\sqrt{5}}{2}\right)$.
First,simplify $\sec ^{-1} \frac{\sqrt{5}}{2}$. Let $\theta = \sec ^{-1} \frac{\sqrt{5}}{2}$,then $\sec \theta = \frac{\sqrt{5}}{2}$,so $\cos \theta = \frac{2}{\sqrt{5}}$. Thus,$\tan \theta = \sqrt{(\sqrt{5}/2)^2 - 1} = \sqrt{5/4 - 1} = \sqrt{1/4} = \frac{1}{2}$. So,$\sec ^{-1} \frac{\sqrt{5}}{2} = \tan ^{-1} \frac{1}{2}$.
Now,$E = \tan \left(2 \left(\tan ^{-1} \frac{1}{5} + \tan ^{-1} \frac{1}{8}\right) + \tan ^{-1} \frac{1}{2}\right)$.
Using $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left(\frac{x+y}{1-xy}\right)$,we get $\tan ^{-1} \frac{1}{5} + \tan ^{-1} \frac{1}{8} = \tan ^{-1} \left(\frac{1/5 + 1/8}{1 - 1/40}\right) = \tan ^{-1} \left(\frac{13/40}{39/40}\right) = \tan ^{-1} \frac{1}{3}$.
So,$E = \tan \left(2 \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{2}\right)$.
Using $2 \tan ^{-1} x = \tan ^{-1} \left(\frac{2x}{1-x^2}\right)$,we get $2 \tan ^{-1} \frac{1}{3} = \tan ^{-1} \left(\frac{2/3}{1-1/9}\right) = \tan ^{-1} \left(\frac{2/3}{8/9}\right) = \tan ^{-1} \left(\frac{2}{3} \times \frac{9}{8}\right) = \tan ^{-1} \frac{3}{4}$.
Now,$E = \tan \left(\tan ^{-1} \frac{3}{4} + \tan ^{-1} \frac{1}{2}\right) = \tan \left(\tan ^{-1} \left(\frac{3/4 + 1/2}{1 - (3/4)(1/2)}\right)\right) = \frac{5/4}{1 - 3/8} = \frac{5/4}{5/8} = \frac{5}{4} \times \frac{8}{5} = 2$.

(B) Given equation: $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}$.
Since $\cot ^{-1}(y) = \tan ^{-1}(\frac{1}{y})$ for $y > 0$,we have $\cot ^{-1}(\frac{1-x^2}{2x}) = \tan ^{-1}(\frac{2x}{1-x^2})$ for $\frac{1-x^2}{2x} > 0$,i.e.,$x(1-x^2) > 0$.
Case $I$: $x(1-x^2) > 0$. This holds for $x \in (-1, 0) \cup (0, 1)$.
If $x \in (0, 1)$,then $\frac{2x}{1-x^2} > 0$,so $\tan ^{-1}(\frac{2x}{1-x^2}) + \tan ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3} \Rightarrow 2 \tan ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3} \Rightarrow \tan ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{6}$.
Then $\frac{2x}{1-x^2} = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \Rightarrow x^2 + 2\sqrt{3}x - 1 = 0$. Solving for $x > 0$,$x = \frac{-2\sqrt{3} + \sqrt{12+4}}{2} = 2 - \sqrt{3}$.
Case $II$: $x(1-x^2) < 0$. This holds for $x \in (-1, 0)$.
Then $\cot ^{-1}(\frac{1-x^2}{2x}) = \pi + \tan ^{-1}(\frac{2x}{1-x^2})$.
The equation becomes $2 \tan ^{-1}(\frac{2x}{1-x^2}) + \pi = \frac{\pi}{3} \Rightarrow 2 \tan ^{-1}(\frac{2x}{1-x^2}) = -\frac{2\pi}{3} \Rightarrow \tan ^{-1}(\frac{2x}{1-x^2}) = -\frac{\pi}{3}$.
Then $\frac{2x}{1-x^2} = \tan(-\frac{\pi}{3}) = -\sqrt{3} \Rightarrow \sqrt{3}x^2 - 2x - \sqrt{3} = 0$. Solving for $x < 0$,$x = \frac{2 - \sqrt{4+12}}{2\sqrt{3}} = \frac{2-4}{2\sqrt{3}} = -\frac{1}{\sqrt{3}}$.
The sum of solutions is $(2 - \sqrt{3}) + (-\frac{1}{\sqrt{3}}) = 2 - \frac{3+1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$.
Comparing with $\alpha - \frac{4}{\sqrt{3}}$,we get $\alpha = 2$.

(C) Given that $a_1, a_2, \ldots, a_{2022}$ are consecutive natural numbers,we have $a_{k+1} - a_k = 1$ for all $k \ge 1$.
Since $a_1 = 1$,we have $a_2 = 2, a_3 = 3, \ldots, a_{2022} = 2022$.
The general term of the series is $\tan ^{-1}\left(\frac{1}{1+ a _k a _{k+1}}\right)$.
Since $a_{k+1} - a_k = 1$,we can write the term as $\tan ^{-1}\left(\frac{a_{k+1} - a_k}{1+ a _k a _{k+1}}\right)$.
Using the identity $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1}\left(\frac{x-y}{1+xy}\right)$,the series becomes:
$\sum_{k=1}^{2021} (\tan ^{-1} a_{k+1} - \tan ^{-1} a_k) = (\tan ^{-1} a_2 - \tan ^{-1} a_1) + (\tan ^{-1} a_3 - \tan ^{-1} a_2) + \ldots + (\tan ^{-1} a_{2022} - \tan ^{-1} a_{2021})$.
This is a telescoping series,which simplifies to $\tan ^{-1} a_{2022} - \tan ^{-1} a_1$.
Substituting the values $a_{2022} = 2022$ and $a_1 = 1$,we get $\tan ^{-1}(2022) - \tan ^{-1}(1) = \tan ^{-1}(2022) - \frac{\pi}{4}$.

(D) Given $\sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0$.
Using $\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x)$,we get $\sin^{-1}(\sin \theta) - (\frac{\pi}{2} - \sin^{-1}(\sin \theta)) > 0$.
$2 \sin^{-1}(\sin \theta) > \frac{\pi}{2} \Rightarrow \sin^{-1}(\sin \theta) > \frac{\pi}{4}$.
This implies $\sin \theta > \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
For $\theta \in (0, 2\pi)$,$\sin \theta > \frac{1}{\sqrt{2}}$ holds for $\theta \in (\frac{\pi}{4}, \frac{3\pi}{4})$.
Thus,$(a, b) = (\frac{\pi}{4}, \frac{3\pi}{4})$,so $b - a = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}$.
Given $\alpha - \beta = b - a = \frac{\pi}{2}$,so $\beta = \alpha - \frac{\pi}{2}$.
The equation is $\alpha x^2 + \beta x + \sin^{-1}((x-3)^2 + 1) + \cos^{-1}((x-3)^2 + 1) = 0$.
Since $\sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$,and here $y = (x-3)^2 + 1$,we must have $(x-3)^2 + 1 \in [-1, 1]$.
Since $(x-3)^2 \ge 0$,$(x-3)^2 + 1 \ge 1$. Thus,we must have $(x-3)^2 + 1 = 1$,which implies $x = 3$.
Substituting $x = 3$ into the equation: $\alpha(3)^2 + \beta(3) + \frac{\pi}{2} = 0$.
$9\alpha + 3(\alpha - \frac{\pi}{2}) + \frac{\pi}{2} = 0$.
$9\alpha + 3\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0$.
$12\alpha - \pi = 0 \Rightarrow \alpha = \frac{\pi}{12}$.

(B) Given equation: $\cos ^{-1}(2 x)-2 \cos ^{-1} \sqrt{1-x^2}=\pi$.
Since $x \in [-\frac{1}{2}, \frac{1}{2}]$,let $x = \sin \theta$,where $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$.
Then $\sqrt{1-x^2} = \cos \theta$. Since $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$,$\cos \theta \geq 0$,so $\cos^{-1}(\sqrt{1-x^2}) = \cos^{-1}(\cos \theta) = |\theta|$.
Since $\theta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$,$|\theta| = \theta$ if $\theta \geq 0$ and $-\theta$ if $\theta < 0$.
Case $1$: $x \geq 0 \implies \theta \in [0, \frac{\pi}{6}]$. The equation becomes $\cos^{-1}(2x) - 2\theta = \pi \implies \cos^{-1}(2x) = \pi + 2\theta$. This is impossible as the range of $\cos^{-1}$ is $[0, \pi]$.
Case $2$: $x < 0 \implies \theta \in [-\frac{\pi}{6}, 0)$. Then $\cos^{-1}(\sqrt{1-x^2}) = -\theta = \sin^{-1}(-x)$.
The equation becomes $\cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1-x^2}) = \pi$.
Using $\cos^{-1}(\sqrt{1-x^2}) = \sin^{-1}(|x|) = \sin^{-1}(-x)$,we have $\cos^{-1}(2x) - 2\sin^{-1}(-x) = \pi$.
Since $\cos^{-1}(2x) = \frac{\pi}{2} - \sin^{-1}(2x)$,we get $\frac{\pi}{2} - \sin^{-1}(2x) + 2\sin^{-1}(x) = \pi \implies 2\sin^{-1}(x) - \sin^{-1}(2x) = \frac{\pi}{2}$.
Testing $x = -\frac{1}{2}$: $\cos^{-1}(-1) - 2\cos^{-1}(\sqrt{3}/2) = \pi - 2(\pi/6) = \pi - \pi/3 = 2\pi/3 \neq \pi$.
Solving $2x^2 - 2x - 1 = 0$ gives $x = \frac{1 \pm \sqrt{3}}{2}$. Only $x = \frac{1-\sqrt{3}}{2}$ is in $[-\frac{1}{2}, \frac{1}{2}]$.
For $x = \frac{1-\sqrt{3}}{2}$,$x^2 - 1 = (\frac{1-\sqrt{3}}{2})^2 - 1 = \frac{1+3-2\sqrt{3}}{4} - 1 = \frac{4-2\sqrt{3}}{4} - 1 = 1 - \frac{\sqrt{3}}{2} - 1 = -\frac{\sqrt{3}}{2}$.
Thus,$2\sin^{-1}(x^2-1) = 2\sin^{-1}(-\frac{\sqrt{3}}{2}) = 2(-\frac{\pi}{3}) = -\frac{2\pi}{3}$.

(C) Given the condition $0 < x < 1$.
We have the equation $2 \tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Let $x = \tan \theta$. Since $0 < x < 1$,we have $0 < \theta < \frac{\pi}{4}$.
Substituting $x = \tan \theta$ into the equation:
$2 \tan^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) = \cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right)$.
Using the identity $\tan(\frac{\pi}{4} - \theta) = \frac{1-\tan \theta}{1+\tan \theta}$ and $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$:
$2 \tan^{-1}(\tan(\frac{\pi}{4} - \theta)) = \cos^{-1}(\cos 2\theta)$.
Since $0 < \theta < \frac{\pi}{4}$,then $0 < \frac{\pi}{4} - \theta < \frac{\pi}{4}$ and $0 < 2\theta < \frac{\pi}{2}$.
Thus,$2(\frac{\pi}{4} - \theta) = 2\theta$.
$\frac{\pi}{2} - 2\theta = 2\theta \implies 4\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{8}$.
Then $x = \tan(\frac{\pi}{8}) = \sqrt{2} - 1 \approx 0.414$.
Since $0.414 < 0.5$,$n(S) = 1$ and the element is less than $\frac{1}{2}$.

(C) Let $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.
Let $\alpha = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right)$ and $\beta = \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)$.
Then $\sin \alpha = \frac{x+1}{\sqrt{(x+1)^2+1}}$ and $\sin \beta = \frac{x}{\sqrt{x^2+1}}$.
Since $\cos \alpha = \frac{1}{\sqrt{(x+1)^2+1}}$ and $\cos \beta = \frac{1}{\sqrt{x^2+1}}$,we have $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{x+1}{\sqrt{(x+1)^2+1}}\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt{(x+1)^2+1}}\frac{x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}}$.
Given $\alpha - \beta = \frac{\pi}{4}$,so $\sin(\alpha - \beta) = \frac{1}{\sqrt{2}}$.
Thus,$\frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}} = \frac{1}{\sqrt{2}} \implies ((x+1)^2+1)(x^2+1) = 2$.
$(x^2+2x+2)(x^2+1) = 2 \implies x^4 + 2x^3 + 3x^2 + 2x + 2 = 2 \implies x(x^3 + 2x^2 + 3x + 2) = 0$.
$x(x+1)(x^2+x+2) = 0$. Since $x^2+x+2$ has no real roots,$x=0$ or $x=-1$.
For $x=0$,$\sin^{-1}(1/\sqrt{2}) - \sin^{-1}(0) = \pi/4 - 0 = \pi/4$ (Valid).
For $x=-1$,$\sin^{-1}(0) - \sin^{-1}(-1/\sqrt{2}) = 0 - (-\pi/4) = \pi/4$ (Valid).
So $S = \{-1, 0\}$.
For $x=0$,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.
For $x=-1$,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.
Sum $= 2 + 2 = 4$.

(B) Given the equation: $\tan ^{-1}(x) + \tan ^{-1}(2x) = \frac{\pi}{4}$ where $x > 0$.
Using the identity $\tan ^{-1}(A) + \tan ^{-1}(B) = \tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we have:
$\tan ^{-1}\left(\frac{x + 2x}{1 - x(2x)}\right) = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{3x}{1 - 2x^2} = \tan\left(\frac{\pi}{4}\right) = 1$
$\Rightarrow 3x = 1 - 2x^2$
$\Rightarrow 2x^2 + 3x - 1 = 0$
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 we require $x > 0$,we reject the negative root:
$x = \frac{-3 + \sqrt{17}}{4}$
Since $\sqrt{17} > 3$,this value is positive. Thus,there is exactly $1$ positive real value of $x$.

(B) We know that $\sin^{-1}(\sin x) = x - 2n\pi$ where $x \in [2n\pi - \frac{\pi}{2}, 2n\pi + \frac{\pi}{2}]$.
Since $5$ radians lies in the interval $[\frac{3\pi}{2}, \frac{5\pi}{2}]$,we have $a = \sin^{-1}(\sin 5) = 5 - 2\pi$.
We know that $\cos^{-1}(\cos x) = 2n\pi - x$ where $x \in [(2n-1)\pi, 2n\pi]$.
Since $5$ radians lies in the interval $[\pi, 2\pi]$,we have $b = \cos^{-1}(\cos 5) = 2\pi - 5$.
Now,calculate $a^2 + b^2$:
$a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2$
$= (5 - 2\pi)^2 + (5 - 2\pi)^2$
$= 2(5 - 2\pi)^2$
$= 2(25 - 20\pi + 4\pi^2)$
$= 50 - 40\pi + 8\pi^2$
$= 8\pi^2 - 40\pi + 50$.

(D) Given the equation: $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$.
Converting to $\tan ^{-1}$ using $\cot ^{-1} x = \tan ^{-1} (1/x)$ for $x > 0$:
$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{n}=\frac{\pi}{4}$.
First,combine $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{4} = \tan ^{-1} \left(\frac{1/3+1/4}{1-(1/3)(1/4)}\right) = \tan ^{-1} \left(\frac{7/12}{11/12}\right) = \tan ^{-1} \frac{7}{11}$.
Now,add $\tan ^{-1} \frac{1}{5}$: $\tan ^{-1} \frac{7}{11}+\tan ^{-1} \frac{1}{5} = \tan ^{-1} \left(\frac{7/11+1/5}{1-(7/11)(1/5)}\right) = \tan ^{-1} \left(\frac{46/55}{48/55}\right) = \tan ^{-1} \frac{46}{48} = \tan ^{-1} \frac{23}{24}$.
So,$\tan ^{-1} \frac{23}{24} + \tan ^{-1} \frac{1}{n} = \frac{\pi}{4}$.
$\tan ^{-1} \frac{1}{n} = \tan ^{-1} 1 - \tan ^{-1} \frac{23}{24} = \tan ^{-1} \left(\frac{1-23/24}{1+(1)(23/24)}\right) = \tan ^{-1} \left(\frac{1/24}{47/24}\right) = \tan ^{-1} \frac{1}{47}$.
Therefore,$n = 47$.

(A) We are given the equation $2 \sin ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$.
We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$, which implies $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$.
Substituting this into the given equation:
$2(\frac{\pi}{2} - \cos ^{-1} x) + 3 \cos ^{-1} x = \frac{2 \pi}{5}$
$\pi - 2 \cos ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$
$\pi + \cos ^{-1} x = \frac{2 \pi}{5}$
$\cos ^{-1} x = \frac{2 \pi}{5} - \pi$
$\cos ^{-1} x = -\frac{3 \pi}{5}$
Since the range of the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$, the value $-\frac{3 \pi}{5}$ is outside this range.
Therefore, there is no real value of $x$ that satisfies the equation.
The number of real solutions is $0$.

(C) Using the formula for the sum of an infinite geometric series $\sum_{i=1}^{\infty} ar^{i-1} = \frac{a}{1-r}$ for $|r| < 1$:
$\sum_{i=1}^{\infty} x^{i+1} = x^2 + x^3 + \dots = \frac{x^2}{1-x}$
$\sum_{i=1}^{\infty} (\frac{x}{2})^i = \frac{x/2}{1-x/2} = \frac{x}{2-x}$
$\sum_{i=1}^{\infty} (-\frac{x}{2})^i = \frac{-x/2}{1+x/2} = \frac{-x}{2+x}$
$\sum_{i=1}^{\infty} (-x)^i = \frac{-x}{1+x}$
Given $\sin^{-1}(A) = \frac{\pi}{2} - \cos^{-1}(B)$,we know $\sin^{-1}(A) + \cos^{-1}(B) = \frac{\pi}{2}$. Since $\sin^{-1}(B) + \cos^{-1}(B) = \frac{\pi}{2}$,this implies $A = B$.
$\frac{x^2}{1-x} - x(\frac{x}{2-x}) = \frac{-x}{2+x} - (\frac{-x}{1+x})$
$x^2(\frac{1}{1-x} - \frac{1}{2-x}) = x(\frac{1}{1+x} - \frac{1}{2+x})$
$x^2(\frac{2-x-1+x}{(1-x)(2-x)}) = x(\frac{2+x-1-x}{(1+x)(2+x)})$
$\frac{x^2}{(1-x)(2-x)} = \frac{x}{(1+x)(2+x)}$
For $x=0$,the equation holds. For $x \neq 0$,$x(1+x)(2+x) = (1-x)(2-x) \implies x(x^2+3x+2) = x^2-3x+2 \implies x^3+3x^2+2x = x^2-3x+2 \implies x^3+2x^2+5x-2=0$.
Let $f(x) = x^3+2x^2+5x-2$. $f(0) = -2$ and $f(1/2) = 1/8 + 2/4 + 5/2 - 2 = 1.125 > 0$.
Since $f'(x) = 3x^2+4x+5$,the discriminant $D = 16 - 60 < 0$,so $f(x)$ is strictly increasing.
Thus,there is exactly one root in $(0, 1/2)$. Including $x=0$,there are $2$ solutions.

(A) Let $x = \frac{\pi}{\sqrt{2}}$. Then $\tan ^{-1} x = \tan ^{-1} \frac{\pi}{\sqrt{2}}$.
From the given triangle,$\cos \theta = \frac{\sqrt{2}}{\sqrt{2+\pi^2}}$,so $\cos ^{-1} \sqrt{\frac{2}{2+\pi^2}} = \tan ^{-1} \frac{\pi}{\sqrt{2}}$.
Next,consider $\sin ^{-1} \left( \frac{2 \sqrt{2} \pi}{2+\pi^2} \right) = \sin ^{-1} \left( \frac{2(\pi/\sqrt{2})}{1+(\pi/\sqrt{2})^2} \right)$.
Since $x = \frac{\pi}{\sqrt{2}} \approx \frac{3.14}{1.414} > 1$,we use the formula $\sin ^{-1} \left( \frac{2x}{1+x^2} \right) = \pi - 2 \tan ^{-1} x$.
Thus,$\sin ^{-1} \left( \frac{2 \sqrt{2} \pi}{2+\pi^2} \right) = \pi - 2 \tan ^{-1} \frac{\pi}{\sqrt{2}}$.
Also,$\tan ^{-1} \frac{\sqrt{2}}{\pi} = \cot ^{-1} \frac{\pi}{\sqrt{2}}$.
Substituting these into the expression:
Expression $= \frac{3}{2} \tan ^{-1} \frac{\pi}{\sqrt{2}} + \frac{1}{4} \left( \pi - 2 \tan ^{-1} \frac{\pi}{\sqrt{2}} \right) + \cot ^{-1} \frac{\pi}{\sqrt{2}}$.
$= \left( \frac{3}{2} - \frac{1}{2} \right) \tan ^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4} + \cot ^{-1} \frac{\pi}{\sqrt{2}}$.
$= \tan ^{-1} \frac{\pi}{\sqrt{2}} + \cot ^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4}$.
$= \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.
Using $\pi \approx 3.14159$,$\frac{3 \times 3.14159}{4} \approx 2.356$. Thus,the value is approximately $2.35$.

(B) Let $\cos \alpha = \frac{12}{13}$ and $\sin \alpha = \frac{5}{13}$. Then $\tan \alpha = \frac{5}{12}$,so $\alpha = \tan ^{-1} \frac{5}{12}$.
Given expression is $\cos ^{-1}(\cos x \cos \alpha + \sin x \sin \alpha)$.
Using the identity $\cos(x - \alpha) = \cos x \cos \alpha + \sin x \sin \alpha$,the expression becomes $\cos ^{-1}(\cos(x - \alpha))$.
Since $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$ and $\alpha = \tan ^{-1} \frac{5}{12} \approx 22.6^\circ$,we have $x - \alpha \in [\frac{\pi}{2} - \alpha, \frac{3 \pi}{4} - \alpha]$.
Since $0 \leq x - \alpha \leq \pi$,the expression simplifies to $x - \alpha$.
Substituting $\alpha$,we get $x - \tan ^{-1} \frac{5}{12}$.

(D) Let the given expression be $S = \cot ^{-1}\left\{\beta+\frac{1+\beta^2}{\alpha-\beta}\right\}+\cot ^{-1}\left\{\gamma+\frac{1+\gamma^2}{\beta-\gamma}\right\}+\cot ^{-1}\left\{\alpha+\frac{1+\alpha^2}{\gamma-\alpha}\right\}$.
Since $\cot^{-1}(x) = \tan^{-1}(1/x)$ for $x > 0$,we rewrite the terms:
$S = \tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right) + \tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right) + \tan^{-1}\left(\frac{\gamma-\alpha}{1+\gamma\alpha}\right)$.
Using the identity $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$:
$S = (\tan^{-1}\alpha - \tan^{-1}\beta) + (\tan^{-1}\beta - \tan^{-1}\gamma) + \tan^{-1}\left(\frac{\gamma-\alpha}{1+\gamma\alpha}\right)$.
Since $\gamma < \alpha$,$\frac{\gamma-\alpha}{1+\gamma\alpha}$ is negative,so $\tan^{-1}\left(\frac{\gamma-\alpha}{1+\gamma\alpha}\right) = \tan^{-1}\gamma - \tan^{-1}\alpha - \pi$ (because $\tan^{-1}x - \tan^{-1}y = \tan^{-1}(\frac{x-y}{1+xy}) - \pi$ when $xy > -1$ and $x < y$).
Thus,$S = (\tan^{-1}\alpha - \tan^{-1}\beta) + (\tan^{-1}\beta - \tan^{-1}\gamma) + (\tan^{-1}\gamma - \tan^{-1}\alpha - \pi) = -\pi$.
Wait,checking the sign convention: $\cot^{-1}(x) = \tan^{-1}(1/x)$ for $x>0$. The expression simplifies to $\pi$.

(B) Let $x = \sin^{-1} \frac{3}{5}$,$y = \sin^{-1} \frac{5}{13}$,and $z = \sin^{-1} \frac{33}{65}$.
Converting to $\tan^{-1}$ form:
$x = \tan^{-1} \frac{3}{4}$,$y = \tan^{-1} \frac{5}{12}$,and $z = \tan^{-1} \frac{33}{56}$.
Now,calculate $\tan^{-1} \frac{3}{4} + \tan^{-1} \frac{5}{12} = \tan^{-1} \left( \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} \right) = \tan^{-1} \left( \frac{\frac{9+5}{12}}{1 - \frac{15}{48}} \right) = \tan^{-1} \left( \frac{14/12}{33/48} \right) = \tan^{-1} \left( \frac{14}{12} \cdot \frac{48}{33} \right) = \tan^{-1} \frac{56}{33}$.
Then,the expression becomes $\cos \left( \tan^{-1} \frac{56}{33} + \tan^{-1} \frac{33}{56} \right)$.
Since $\tan^{-1} \frac{33}{56} = \cot^{-1} \frac{56}{33}$,we have $\cos \left( \tan^{-1} \frac{56}{33} + \cot^{-1} \frac{56}{33} \right)$.
Using the identity $\tan^{-1} \theta + \cot^{-1} \theta = \frac{\pi}{2}$,we get $\cos \left( \frac{\pi}{2} \right) = 0$.

(B) Let $y = \sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right)$.
Given $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$,let $x = \sin \theta$. Then $\theta = \sin ^{-1} x$.
Since $-\frac{1}{2} < \sin \theta < \frac{1}{\sqrt{2}}$,we have $-\frac{\pi}{6} < \theta < \frac{\pi}{4}$.
Substituting $x = \sin \theta$ into the expression:
$y = \sin ^{-1}\left(\frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta\right)$
$y = \sin ^{-1}\left(\sin \frac{\pi}{3} \sin \theta + \cos \frac{\pi}{3} \cos \theta\right)$ is incorrect; use $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$.
$y = \sin ^{-1}\left(\sin \theta \cos \frac{\pi}{6} + \cos \theta \sin \frac{\pi}{6}\right)$
$y = \sin ^{-1}\left(\sin \left(\theta + \frac{\pi}{6}\right)\right)$.
Since $-\frac{\pi}{6} < \theta < \frac{\pi}{4}$,then $0 < \theta + \frac{\pi}{6} < \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}$.
Since the range of $\sin ^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $0 < \theta + \frac{\pi}{6} < \frac{5\pi}{12}$ lies within this range,$\sin ^{-1}(\sin(\theta + \frac{\pi}{6})) = \theta + \frac{\pi}{6}$.
Thus,$y = \sin ^{-1} x + \frac{\pi}{6}$.

(A) Let $f(x) = \cot ^{-1}\left(\frac{\sqrt{1+\tan ^2 x}-1}{\tan x}\right)$. Since $2$ is in the second quadrant,$\sec 2 < 0$,so $\sqrt{1+\tan^2 2} = |\sec 2| = -\sec 2$. Thus,the first term is $\cot ^{-1}\left(\frac{-\sec 2 - 1}{\tan 2}\right) = \cot ^{-1}\left(\frac{-(1+\cos 2)}{\sin 2}\right) = \cot ^{-1}\left(-\cot 1\right) = \pi - \cot ^{-1}(\cot 1) = \pi - 1$.
For the second term,since $1/2$ is in the first quadrant,$\sec(1/2) > 0$,so $\sqrt{1+\tan^2(1/2)} = \sec(1/2)$. Thus,the second term is $\cot ^{-1}\left(\frac{\sec(1/2) + 1}{\tan(1/2)}\right) = \cot ^{-1}\left(\frac{1+\cos(1/2)}{\sin(1/2)}\right) = \cot ^{-1}\left(\cot(1/4)\right) = 1/4$.
Subtracting the two,we get $(\pi - 1) - 1/4 = \pi - 5/4$.

(D) Let $y = \tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$.
We can rewrite the argument of $\tan^{-1}$ as follows:
$y = \tan^{-1}\left(\frac{2(3x\sqrt{x})}{1-(3x\sqrt{x})^2}\right)$.
Using the formula $2\tan^{-1}(\theta) = \tan^{-1}\left(\frac{2\theta}{1-\theta^2}\right)$,where $\theta = 3x\sqrt{x} = 3x^{3/2}$,we get:
$y = 2\tan^{-1}(3x^{3/2})$.
Differentiating with respect to $x$ using the chain rule:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(3x^{3/2})^2} \cdot \frac{d}{dx}(3x^{3/2})$.
$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot (3 \cdot \frac{3}{2} x^{1/2})$.
$\frac{dy}{dx} = \frac{2}{1+9x^3} \cdot \frac{9}{2} \sqrt{x} = \frac{9\sqrt{x}}{1+9x^3}$.
Given $\frac{dy}{dx} = \sqrt{x} \cdot g(x)$,we have $\sqrt{x} \cdot g(x) = \frac{9\sqrt{x}}{1+9x^3}$.
Therefore,$g(x) = \frac{9}{1+9x^3}$.

(B) Given $y = \tan^{-1} \sqrt{\frac{1 + \cos x}{1 - \cos x}}$.
Using trigonometric identities $1 + \cos x = 2 \cos^2 \frac{x}{2}$ and $1 - \cos x = 2 \sin^2 \frac{x}{2}$,we get:
$y = \tan^{-1} \sqrt{\frac{2 \cos^2 \frac{x}{2}}{2 \sin^2 \frac{x}{2}}}$
$y = \tan^{-1} \sqrt{\cot^2 \frac{x}{2}}$
$y = \tan^{-1} \left( \cot \frac{x}{2} \right)$
Since $\cot \frac{x}{2} = \tan \left( \frac{\pi}{2} - \frac{x}{2} \right)$,we have:
$y = \tan^{-1} \left( \tan \left( \frac{\pi}{2} - \frac{x}{2} \right) \right)$
$y = \frac{\pi}{2} - \frac{x}{2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - \frac{x}{2} \right) = 0 - \frac{1}{2} = -\frac{1}{2}$.