Properties of ITF Questions in English

(C) Given $f(n) = \tan \left[\sum_{k=1}^{n} \tan ^{-1} \frac{1}{1+k(k+1)}\right]$.
We know that $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \left(\frac{x-y}{1+xy}\right)$.
We can write the general term as $\tan ^{-1} \left(\frac{(k+1)-k}{1+k(k+1)}\right) = \tan ^{-1}(k+1) - \tan ^{-1}(k)$.
Thus, the sum becomes a telescoping series:
$S_n = (\tan ^{-1}(2) - \tan ^{-1}(1)) + (\tan ^{-1}(3) - \tan ^{-1}(2)) + \ldots + (\tan ^{-1}(n+1) - \tan ^{-1}(n))$.
All intermediate terms cancel out, leaving $S_n = \tan ^{-1}(n+1) - \tan ^{-1}(1)$.
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left(\frac{A-B}{1+AB}\right)$:
$S_n = \tan ^{-1} \left(\frac{(n+1)-1}{1+(n+1)(1)}\right) = \tan ^{-1} \left(\frac{n}{n+2}\right)$.
Therefore, $f(n) = \tan \left[\tan ^{-1} \left(\frac{n}{n+2}\right)\right] = \frac{n}{n+2}$.
For $n = 2021$, $f(2021) = \frac{2021}{2021+2} = \frac{2021}{2023}$.

(C) We have, $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right) = \tan ^{-1} \alpha$.
Using the identity $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1}\left(\frac{x-y}{1+xy}\right)$, we can rewrite the term inside the summation:
$\frac{1}{n^2+3n+3} = \frac{(n+2)-(n+1)}{1+(n+2)(n+1)}$.
Thus, the summation becomes:
$\sum_{n=1}^k (\tan ^{-1}(n+2) - \tan ^{-1}(n+1)) = \tan ^{-1} \alpha$.
Expanding the sum:
$(\tan ^{-1} 3 - \tan ^{-1} 2) + (\tan ^{-1} 4 - \tan ^{-1} 3) + \dots + (\tan ^{-1}(k+2) - \tan ^{-1}(k+1)) = \tan ^{-1} \alpha$.
This is a telescoping series, so all intermediate terms cancel out:
$\tan ^{-1}(k+2) - \tan ^{-1} 2 = \tan ^{-1} \alpha$.
Applying the formula again:
$\tan ^{-1}\left(\frac{(k+2)-2}{1+(k+2)(2)}\right) = \tan ^{-1} \alpha$.
$\tan ^{-1}\left(\frac{k}{1+2k+4}\right) = \tan ^{-1} \alpha$.
Therefore, $\alpha = \frac{k}{2k+5}$.

(A) The given equation is $2 \tan^{-1} 2x = \sin^{-1} \left( \frac{4x}{1+4x^2} \right)$.
We know the identity $\sin^{-1} \left( \frac{2\theta}{1+\theta^2} \right) = 2 \tan^{-1} \theta$, which holds true when $-1 \leq \theta \leq 1$.
Here, let $\theta = 2x$. The equation becomes $2 \tan^{-1} 2x = \sin^{-1} \left( \frac{2(2x)}{1+(2x)^2} \right)$.
This identity is valid if and only if $-1 \leq 2x \leq 1$.
Dividing by $2$, we get $-\frac{1}{2} \leq x \leq \frac{1}{2}$.
Thus, the values of $x$ lie in the interval $[-\frac{1}{2}, \frac{1}{2}]$.

(D) Given, $y=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)+\tan ^{-1}\left(\frac{4 x-4 x^3}{1-6 x^2+x^4}\right)$.
Substitute $x=\tan \theta$, then $\theta = \tan^{-1} x$.
The expression becomes:
$y = \tan^{-1}(\tan 3\theta) + \tan^{-1}(\tan 4\theta)$.
This simplifies to $y = 3\theta + 4\theta = 7\theta$.
Substituting back, $y = 7 \tan^{-1} x$.
Differentiating with respect to $x$:
$\frac{d y}{d x} = 7 \times \frac{1}{1+x^2} = \frac{7}{1+x^2}$.

(D) We know that the range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\sin ^{-1} x = 2 \sin ^{-1} 2a$, the value of $2 \sin ^{-1} 2a$ must lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$\Rightarrow -\frac{\pi}{2} \leq 2 \sin ^{-1} 2a \leq \frac{\pi}{2}$
$\Rightarrow -\frac{\pi}{4} \leq \sin ^{-1} 2a \leq \frac{\pi}{4}$
Taking sine on all sides, we get:
$\sin(-\frac{\pi}{4}) \leq 2a \leq \sin(\frac{\pi}{4})$
$\Rightarrow -\frac{1}{\sqrt{2}} \leq 2a \leq \frac{1}{\sqrt{2}}$
Dividing by $2$, we get:
$-\frac{1}{2\sqrt{2}} \leq a \leq \frac{1}{2\sqrt{2}}$
This is equivalent to $|a| \leq \frac{1}{2\sqrt{2}}$.

(B) The $n^{\text{th}}$ term of the series is $t_{n} = \cot^{-1}(2n^2)$.
Using the identity $\cot^{-1} x = \tan^{-1} \frac{1}{x}$, we have $t_{n} = \tan^{-1} \frac{1}{2n^2}$.
We can rewrite this as $t_{n} = \tan^{-1} \frac{2}{4n^2} = \tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$.
Using the formula $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy}$, we get $t_{n} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$.
The sum $S_{n} = \sum_{k=1}^{n} t_{k} = (\tan^{-1} 3 - \tan^{-1} 1) + (\tan^{-1} 5 - \tan^{-1} 3) + \dots + (\tan^{-1}(2n+1) - \tan^{-1}(2n-1))$.
This is a telescoping series, so $S_{n} = \tan^{-1}(2n+1) - \tan^{-1} 1$.
Taking the limit as $n \rightarrow \infty$, $\lim_{n \rightarrow \infty} S_{n} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(C) We know that the range of $\sin ^{-1} \theta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given that $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$.
Since the maximum value of each term $\sin ^{-1} x, \sin ^{-1} y, \sin ^{-1} z$ is $\frac{\pi}{2}$, the sum can be $\frac{3 \pi}{2}$ only if $\sin ^{-1} x = \frac{\pi}{2}$, $\sin ^{-1} y = \frac{\pi}{2}$, 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:
$x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} = (1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}}$
$= 1+1+1-\frac{1}{1 \times 1 \times 1}$
$= 3-1 = 2$.

(D) We are given the inequation $\cos ^{-1} x < \sin ^{-1} x$.
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 inequation, we get:
$\frac{\pi}{2} - \sin ^{-1} x < \sin ^{-1} x$
$\frac{\pi}{2} < 2 \sin ^{-1} x$
$\sin ^{-1} x > \frac{\pi}{4}$
Taking the sine of both sides (since $\sin x$ is an increasing function in its domain):
$x > \sin\left(\frac{\pi}{4}\right)$
$x > \frac{1}{\sqrt{2}}$
Since the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $[-1, 1]$, we must have $x \le 1$.
Therefore, the solution set is $x \in \left(\frac{1}{\sqrt{2}}, 1\right]$.

(A) Given, $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} \quad ...(i)$
We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$.
Therefore, $\operatorname{cosec}^{-1}\left(\frac{13}{12}\right) = \sin^{-1}\left(\frac{12}{13}\right)$.
Substituting this into equation $(i)$, we get:
$\sin ^{-1}\left(\frac{x}{13}\right)+\sin ^{-1}\left(\frac{12}{13}\right)=\frac{\pi}{2}$.
We know the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$.
Also, $\sin^{-1}(\frac{12}{13}) = \cos^{-1}(\sqrt{1 - (\frac{12}{13})^2}) = \cos^{-1}(\sqrt{1 - \frac{144}{169}}) = \cos^{-1}(\sqrt{\frac{25}{169}}) = \cos^{-1}(\frac{5}{13})$.
So, $\sin^{-1}(\frac{x}{13}) + \cos^{-1}(\frac{5}{13}) = \frac{\pi}{2}$.
Comparing this with $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$, we must have $\frac{x}{13} = \frac{5}{13}$.
Thus, $x = 5$.

(C) Given that $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$.
We know that the range of $\cos ^{-1} x$ is $[0, \pi]$.
Since the sum of three values, each at most $\pi$, is $3 \pi$, each term must be equal to $\pi$.
Therefore, $\cos ^{-1} \alpha = \pi$, $\cos ^{-1} \beta = \pi$, and $\cos ^{-1} \gamma = \pi$.
This implies $\alpha = \cos(\pi) = -1$, $\beta = \cos(\pi) = -1$, and $\gamma = \cos(\pi) = -1$.
Now, we calculate the expression $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$.
Substituting $\alpha = -1$, $\beta = -1$, and $\gamma = -1$:
$(-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1) = (-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6$.

(D) Given expression: $2 \cot ^{-1} \frac{1}{2} - \cot ^{-1} \frac{4}{3}$
Using the property $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$:
$= 2 \tan ^{-1} 2 - \tan ^{-1} \frac{3}{4}$
Using the formula $2 \tan ^{-1} x = \pi + \tan ^{-1} \frac{2x}{1-x^2}$ for $x > 1$:
$= \pi + \tan ^{-1} \frac{2(2)}{1-2^2} - \tan ^{-1} \frac{3}{4}$
$= \pi + \tan ^{-1} \frac{4}{-3} - \tan ^{-1} \frac{3}{4}$
$= \pi - \tan ^{-1} \frac{4}{3} - \tan ^{-1} \frac{3}{4}$
$= \pi - (\tan ^{-1} \frac{4}{3} + \tan ^{-1} \frac{3}{4})$
Using the property $\tan ^{-1} x + \tan ^{-1} \frac{1}{x} = \frac{\pi}{2}$ for $x > 0$:
$= \pi - \frac{\pi}{2} = \frac{\pi}{2}$

(B) Let $f(x) = (\sin^{-1}x)^{2} + (\cos^{-1}x)^{2}$.
Since $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$, we have:
$f(x) = (\sin^{-1}x)^{2} + (\frac{\pi}{2} - \sin^{-1}x)^{2}$
$f(x) = (\sin^{-1}x)^{2} + \frac{\pi^{2}}{4} - \pi \sin^{-1}x + (\sin^{-1}x)^{2}$
$f(x) = 2(\sin^{-1}x)^{2} - \pi \sin^{-1}x + \frac{\pi^{2}}{4}$
$f(x) = 2[(\sin^{-1}x)^{2} - \frac{\pi}{2} \sin^{-1}x] + \frac{\pi^{2}}{4}$
$f(x) = 2[(\sin^{-1}x - \frac{\pi}{4})^{2} - \frac{\pi^{2}}{16}] + \frac{\pi^{2}}{4}$
$f(x) = 2(\sin^{-1}x - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8}$.
Given $x \in [-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}]$, the range of $\sin^{-1}x$ is $[-\frac{\pi}{3}, \frac{\pi}{4}]$.
To maximize $f(x)$, we choose the value of $\sin^{-1}x$ furthest from $\frac{\pi}{4}$, which is $-\frac{\pi}{3}$.
Max value $= 2(-\frac{\pi}{3} - \frac{\pi}{4})^{2} + \frac{\pi^{2}}{8} = 2(-\frac{7\pi}{12})^{2} + \frac{\pi^{2}}{8} = 2(\frac{49\pi^{2}}{144}) + \frac{\pi^{2}}{8} = \frac{49\pi^{2}}{72} + \frac{9\pi^{2}}{72} = \frac{58\pi^{2}}{72} = \frac{29\pi^{2}}{36}$.
Thus, $m = 29$ and $n = 36$. Since $\gcd(29, 36) = 1$, $m+n = 29 + 36 = 65$.

(C) Given equation: $\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$.
Using the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$, we have:
$\tan^{-1}\left(\frac{4x+6x}{1-(4x)(6x)}\right) = \frac{\pi}{6}$.
$\frac{10x}{1-24x^2} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.
$10\sqrt{3}x = 1 - 24x^2 \implies 24x^2 + 10\sqrt{3}x - 1 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}$.
The range is $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, where $\frac{1}{2\sqrt{6}} \approx 0.204$.
$x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} \approx \frac{-8.66 + 9.95}{24} \approx 0.054$ (In range).
$x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} \approx \frac{-8.66 - 9.95}{24} \approx -0.775$ (Not in range).
Only one solution satisfies the condition.

(A) Let $x = \sin\theta$. Since $x \in [-1/2, 1/2]$, we have $\theta \in [-\pi/6, \pi/6]$.
The expression becomes $y = 3\sin^{-1}(\sin\theta) + \sin^{-1}(\sin(3\theta))$.
Since $\theta \in [-\pi/6, \pi/6]$, we have $3\theta \in [-\pi/2, \pi/2]$.
Therefore, $\sin^{-1}(\sin\theta) = \theta$ and $\sin^{-1}(\sin(3\theta)) = 3\theta$.
Substituting these, we get $y = 3\theta + 3\theta = 6\theta$.
Given $\theta \in [-\pi/6, \pi/6]$, multiplying by $6$ gives $6\theta \in [-\pi, \pi]$.
Thus, $y \in [-\pi, \pi]$.

(B) Given the equation $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$.
Using the formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, we get $\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)} = \tan\left(\frac{\pi}{4}\right) = 1$.
This simplifies to $2 - (\alpha + \beta) = 1 - (1 - \alpha - \beta + \alpha\beta)$.
$2 - \alpha - \beta = \alpha + \beta - \alpha\beta \Rightarrow 2 = 2(\alpha + \beta) - \alpha\beta$.
Substitute $\beta = \frac{1}{3\alpha}$ into the equation:
$2 = 2(\alpha + \frac{1}{3\alpha}) - \alpha(\frac{1}{3\alpha}) = 2\alpha + \frac{2}{3\alpha} - \frac{1}{3}$.
$2 + \frac{1}{3} = 2\alpha + \frac{2}{3\alpha} \Rightarrow \frac{7}{3} = \frac{6\alpha^2 + 2}{3\alpha}$.
$7\alpha = 6\alpha^2 + 2 \Rightarrow 6\alpha^2 - 7\alpha + 2 = 0$.
Factoring the quadratic: $(2\alpha - 1)(3\alpha - 2) = 0$.
Thus, $\alpha = \frac{1}{2}$ or $\alpha = \frac{2}{3}$.
If $\alpha = \frac{1}{2}$, then $\beta = \frac{1}{3(1/2)} = \frac{2}{3}$.
If $\alpha = \frac{2}{3}$, then $\beta = \frac{1}{3(2/3)} = \frac{1}{2}$.
In both cases, $\alpha + \beta = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}$.
Therefore, $6(\alpha + \beta) = 6 \times \frac{7}{6} = 7$.

(A) We use the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$.
Given the term $\tan^{-1} \left( \frac{2^{p-1}}{1+2^{2p-1}} \right)$, we can rewrite it as $\tan^{-1} \left( \frac{2^p - 2^{p-1}}{1 + 2^p \cdot 2^{p-1}} \right) = \tan^{-1}(2^p) - \tan^{-1}(2^{p-1})$.
Now, sum the telescoping series: $\sum_{p=1}^{11} (\tan^{-1}(2^p) - \tan^{-1}(2^{p-1})) = (\tan^{-1}(2^1) - \tan^{-1}(2^0)) + (\tan^{-1}(2^2) - \tan^{-1}(2^1)) + \dots + (\tan^{-1}(2^{11}) - \tan^{-1}(2^{10}))$.
This simplifies to $\tan^{-1}(2^{11}) - \tan^{-1}(2^0) = \tan^{-1}(2048) - \frac{\pi}{4}$.
Substituting this back into the original equation: $\frac{\pi}{4} + (\tan^{-1}(2048) - \frac{\pi}{4}) = \tan^{-1}(2048)$.
Thus, $\tan^{-1} \alpha = \tan^{-1}(2048)$, which implies $\alpha = 2048$.
Therefore, $\tan \alpha = \tan(2048)$.

(C) We know that $\tan^{-1} x - \tan^{-1} y = \tan^{-1} (\frac{x - y}{1 + xy})$.
Since $a_1, a_2, \dots, a_n$ are in arithmetic progression, $a_{k+1} - a_k = d$.
Thus, each term in the sum can be written as $\tan^{-1} (\frac{a_{k+1} - a_k}{1 + a_k a_{k+1}}) = \tan^{-1} a_{k+1} - \tan^{-1} a_k$.
The given expression becomes $\tan [(\tan^{-1} a_2 - \tan^{-1} a_1) + (\tan^{-1} a_3 - \tan^{-1} a_2) + \dots + (\tan^{-1} a_n - \tan^{-1} a_{n-1})]$.
This is a telescoping sum, which simplifies to $\tan [\tan^{-1} a_n - \tan^{-1} a_1]$.
Using the formula $\tan (\tan^{-1} x - \tan^{-1} y) = \frac{x - y}{1 + xy}$, we get $\frac{a_n - a_1}{1 + a_n a_1}$.

(C) Given $\sec^{-1} \left( \frac{x^2 + y^2}{x^2 - y^2} \right) = 2a$, we have $\frac{x^2 + y^2}{x^2 - y^2} = \sec(2a)$.
Applying componendo and dividendo: $\frac{(x^2 + y^2) + (x^2 - y^2)}{(x^2 + y^2) - (x^2 - y^2)} = \frac{\sec(2a) + 1}{\sec(2a) - 1}$.
$\frac{2x^2}{2y^2} = \frac{\frac{1}{\cos(2a)} + 1}{\frac{1}{\cos(2a)} - 1} = \frac{1 + \cos(2a)}{1 - \cos(2a)} = \frac{2\cos^2(a)}{2\sin^2(a)} = \cot^2(a)$.
So, $\frac{x^2}{y^2} = \cot^2(a) \implies x^2 = y^2 \cot^2(a)$.
Differentiating with respect to $x$: $2x = 2y \frac{dy}{dx} \cot^2(a)$.
$x = y \frac{dy}{dx} \cot^2(a) \implies y \frac{dy}{dx} = x \tan^2(a)$.
Comparing with $y \frac{dy}{dx} = x \cdot f(a)$, we get $f(a) = \tan^2(a)$.
Therefore, $f \left( \frac{2\pi}{3} \right) = \tan^2 \left( \frac{2\pi}{3} \right) = (-\sqrt{3})^2 = 3$.

(D) Step $1$: Use the formula $3 \tan^{-1}(x) = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$.
Step $2$: Substitute $x = 1/2$ into the formula.
Step $3$: Calculate the numerator: $3(1/2) - (1/2)^3 = 3/2 - 1/8 = 12/8 - 1/8 = 11/8$.
Step $4$: Calculate the denominator: $1 - 3(1/2)^2 = 1 - 3/4 = 1/4$.
Step $5$: The expression becomes $\tan^{-1}\left(\frac{11/8}{1/4}\right) = \tan^{-1}\left(\frac{11}{8} \times 4\right) = \tan^{-1}(11/2)$.

(B) Let $\theta = \sin^{-1}(1/5)$, then $\sin \theta = 1/5$.
We need to evaluate $\sin(3\theta)$.
Using the identity $\sin(3\theta) = 3\sin \theta - 4\sin^3 \theta$.
Substitute $\sin \theta = 1/5$ into the identity:
$\sin(3\theta) = 3(1/5) - 4(1/5)^3$
$\sin(3\theta) = 3/5 - 4/125$
$\sin(3\theta) = (75 - 4) / 125 = 71/125$.

(C) Let $x = \tan \theta$. Then $\theta = \tan^{-1} x$.
Using the standard trigonometric substitutions for $|x| \le 1$:
$\sin^{-1}(\frac{2x}{1+x^2}) = 2 \tan^{-1} x = 2\theta$
$\cos^{-1}(\frac{1-x^2}{1+x^2}) = 2 \tan^{-1} x = 2\theta$
$\tan^{-1}(\frac{2x}{1-x^2}) = 2 \tan^{-1} x = 2\theta$
Substituting these into the equation:
$3(2\theta) - 4(2\theta) + 2(2\theta) = \frac{\pi}{3}$
$6\theta - 8\theta + 4\theta = \frac{\pi}{3}$
$2\theta = \frac{\pi}{3}$
$\theta = \frac{\pi}{6}$
Since $x = \tan \theta$, we have $x = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.

(A) Step $1$: Use the property $\tan^{-1}(x) + \tan^{-1}(y) = \pi + \tan^{-1}(\frac{x+y}{1-xy})$ for $xy > 1$.
Step $2$: Combine $\tan^{-1}(3) + \tan^{-1}(5) = \pi + \tan^{-1}(\frac{3+5}{1-15}) = \pi + \tan^{-1}(\frac{8}{-14}) = \pi - \tan^{-1}(4/7)$.
Step $3$: The expression becomes $\tan^{-1}(1) + \pi - \tan^{-1}(4/7) + \tan^{-1}(1/4) = \pi + \tan^{-1}(1) + \tan^{-1}(1/4) - \tan^{-1}(4/7)$.
Step $4$: Use $\tan^{-1}(1) + \tan^{-1}(1/4) = \tan^{-1}(\frac{1+1/4}{1-1/4}) = \tan^{-1}(\frac{5/4}{3/4}) = \tan^{-1}(5/3)$.
Step $5$: Now, $\tan^{-1}(5/3) - \tan^{-1}(4/7) = \tan^{-1}(\frac{5/3 - 4/7}{1 + (5/3)(4/7)}) = \tan^{-1}(\frac{35-12}{21+20}) = \tan^{-1}(23/41)$.
Step $6$: Comparing with $\tan^{-1}(\alpha/2)$, we have $\alpha/2 = 23/41$, so $\alpha = 46/41$.

(C) Let $\alpha = \sin^{-1}(12/13)$, then $\sin \alpha = 12/13$. Thus, $\tan \alpha = 12/5$.
Let $\beta = \cos^{-1}(4/5)$, then $\cos \beta = 4/5$. Thus, $\tan \beta = 3/4$.
Now, $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{12/5 + 3/4}{1 - (12/5)(3/4)} = \frac{(48+15)/20}{1 - 36/20} = \frac{63/20}{-16/20} = -63/16$.
Since $\alpha + \beta$ is in the second quadrant, $\alpha + \beta = \pi - \tan^{-1}(63/16)$.
Therefore, $\alpha + \beta + \tan^{-1}(63/16) = \pi - \tan^{-1}(63/16) + \tan^{-1}(63/16) = \pi$.

(C) We use the identity $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}(\frac{a-b}{1+ab})$.
Note that $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+n(n+1)}$.
Thus, $\tan^{-1}(\frac{1}{n^2+n+1}) = \tan^{-1}(n+1) - \tan^{-1}(n)$.
The sum is $\sum_{n=1}^{2026} (\tan^{-1}(n+1) - \tan^{-1}(n))$.
This is a telescoping sum: $(\tan^{-1}(2) - \tan^{-1}(1)) + (\tan^{-1}(3) - \tan^{-1}(2)) + \dots + (\tan^{-1}(2027) - \tan^{-1}(2026))$.
The sum simplifies to $\tan^{-1}(2027) - \tan^{-1}(1)$.
Using the identity again: $\tan^{-1}(2027) - \tan^{-1}(1) = \tan^{-1}(\frac{2027-1}{1+2027 \cdot 1}) = \tan^{-1}(\frac{2026}{2028}) = \tan^{-1}(\frac{1013}{1014})$.
We are given $\tan^{-1}(1 - \frac{1}{x}) = \tan^{-1}(\frac{1013}{1014})$.
So, $1 - \frac{1}{x} = \frac{1013}{1014} \implies \frac{1}{x} = 1 - \frac{1013}{1014} = \frac{1}{1014}$.
Therefore, $x = 1014$.

(A) Step $1$: Simplify $\sin^{-1}(\sin \frac{7\pi}{6})$. Since $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, $\sin(\frac{7\pi}{6}) = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$. Thus, $\sin^{-1}(\sin \frac{7\pi}{6}) = -\frac{\pi}{6}$.
Step $2$: Simplify $\cos^{-1}(\cos \frac{7\pi}{6})$. Since $\frac{7\pi}{6} = 2\pi - \frac{5\pi}{6}$, $\cos(\frac{7\pi}{6}) = \cos(\frac{5\pi}{6})$. Thus, $\cos^{-1}(\cos \frac{7\pi}{6}) = \frac{5\pi}{6}$.
Step $3$: Simplify $\tan^{-1}(\tan \frac{2\pi}{3})$. Since $\frac{2\pi}{3} = \pi - \frac{\pi}{3}$, $\tan(\frac{2\pi}{3}) = -\tan(\frac{\pi}{3}) = \tan(-\frac{\pi}{3})$. Thus, $\tan^{-1}(\tan \frac{2\pi}{3}) = -\frac{\pi}{3}$.
Step $4$: Sum the values: $-\frac{\pi}{6} + \frac{5\pi}{6} - \frac{\pi}{3} = \frac{4\pi}{6} - \frac{2\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

(B) Step $1$: Find the value of $\cos^{-1}(-\frac{1}{2})$. Since $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$, we have $\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$.
Step $2$: Substitute this into the expression: $\cos(\frac{2\pi}{3} + \frac{\pi}{3})$.
Step $3$: Simplify the angle: $\frac{2\pi}{3} + \frac{\pi}{3} = \frac{3\pi}{3} = \pi$.
Step $4$: Calculate $\cos(\pi) = -1$.

(D) Given expression is $y = \cot^{-1} \left[ \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right]$.
Since $x \in (0, \frac{\pi}{2})$, we have $\frac{x}{2} \in (0, \frac{\pi}{4})$.
Using $1 \pm \sin x = (\cos \frac{x}{2} \pm \sin \frac{x}{2})^2$, we get $\sqrt{1 \pm \sin x} = |\cos \frac{x}{2} \pm \sin \frac{x}{2}|$.
Since $0 < \frac{x}{2} < \frac{\pi}{4}$, $\cos \frac{x}{2} > \sin \frac{x}{2} > 0$.
Thus, $\sqrt{1 + \sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1 - \sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$.
Substituting these, the expression inside $\cot^{-1}$ becomes $\frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) + (\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2}) - (\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{2 \cos \frac{x}{2}}{-2 \sin \frac{x}{2}} = -\cot \frac{x}{2}$.
So, $y = \cot^{-1}(-\cot \frac{x}{2}) = \pi - \cot^{-1}(\cot \frac{x}{2}) = \pi - \frac{x}{2}$.

(C) Step $1$: Simplify $\cos^{-1}(\cos \frac{4\pi}{3})$. Since $\frac{4\pi}{3}$ is not in the range $[0, \pi]$, we write $\cos \frac{4\pi}{3} = \cos(2\pi - \frac{2\pi}{3}) = \cos \frac{2\pi}{3}$. Thus, $\cos^{-1}(\cos \frac{2\pi}{3}) = \frac{2\pi}{3}$.
Step $2$: Simplify $\sin^{-1}(\sin \frac{4\pi}{3})$. Since $\frac{4\pi}{3}$ is not in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we write $\sin \frac{4\pi}{3} = \sin(\pi + \frac{\pi}{3}) = -\sin \frac{\pi}{3} = \sin(-\frac{\pi}{3})$. Thus, $\sin^{-1}(\sin(-\frac{\pi}{3})) = -\frac{\pi}{3}$.
Step $3$: Add the results: $\frac{2\pi}{3} + (-\frac{\pi}{3}) = \frac{\pi}{3}$.

(B) The domain of $\sin^{-1}(x-2)$ is $[1, 3]$, $\cos^{-1}(x)$ is $[-1, 1]$, $\tan^{-1}(x+2)$ is $(-\infty, \infty)$, and $\cot^{-1}(x+4)$ is $(-\infty, \infty)$.
The intersection of these domains is $\{1\}$.
Substituting $x = 1$ into the equation:
$\sin^{-1}(1-2) + \cos^{-1}(1) + \tan^{-1}(1+2) + \cot^{-1}(1+4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$
$-\frac{\pi}{2} + 0 + \tan^{-1}(3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$
$\tan^{-1}(3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k})$
Using $\tan^{-1}(3) = \cot^{-1}(1/3)$, we have $\cot^{-1}(1/3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k})$.
Let $\alpha = \cot^{-1}(1/3)$ and $\beta = \cot^{-1}(5)$. Then $\cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = \frac{(1/3)(5) - 1}{1/3 + 5} = \frac{2/3}{16/3} = \frac{1}{8}$.
So, $\alpha + \beta = \cot^{-1}(1/8) = \tan^{-1}(8) = \sec^{-1}(\sqrt{1+8^2}) = \sec^{-1}(\sqrt{65})$.
Thus, $\sqrt{k} = \sqrt{65}$, so $k = 65$.
We need to find $\cos(2 \csc^{-1}\sqrt{65-1}) = \cos(2 \csc^{-1}(8))$.
Let $\theta = \csc^{-1}(8)$, so $\csc \theta = 8$, which means $\sin \theta = 1/8$.
Then $\cos(2\theta) = 1 - 2\sin^2 \theta = 1 - 2(1/8)^2 = 1 - 2/64 = 1 - 1/32 = \frac{31}{32}$.

(A) Given the equation: $\tan^{-1} (ax) + \tan^{-1} (3x) = \frac{\pi}{4}$.
Using the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$, we get:
$\tan^{-1} \left( \frac{ax + 3x}{1 - 3ax^2} \right) = \frac{\pi}{4}$.
Taking $\tan$ on both sides: $\frac{x(a+3)}{1-3ax^2} = \tan \left( \frac{\pi}{4} \right) = 1$.
Substitute $x = \frac{1}{6}$ into the equation:
$\frac{\frac{1}{6}(a+3)}{1 - 3a(\frac{1}{6})^2} = 1$.
$\frac{\frac{a+3}{6}}{1 - \frac{3a}{36}} = 1 \implies \frac{a+3}{6} = 1 - \frac{a}{12}$.
Multiply by $12$: $2(a+3) = 12 - a$.
$2a + 6 = 12 - a \implies 3a = 6 \implies a = 2$.

(A) Let $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$.
We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.
Substituting this, $f(x) = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$.
Let $u = \sin^{-1} x$. Since $x \in [-1, 1]$, $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
$f(u) = u^2 + (\frac{\pi}{2} - u)^2 = u^2 + \frac{\pi^2}{4} - \pi u + u^2 = 2u^2 - \pi u + \frac{\pi^2}{4}$.
This is a parabola opening upwards. The minimum occurs at $u = -\frac{b}{2a} = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$.
Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the minimum value is $f(\frac{\pi}{4}) = 2(\frac{\pi}{4})^2 - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = 2(\frac{\pi^2}{16}) - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$.

(B) The domain of $f(x)$ is $x \in [-1, 1]$.
We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.
Substituting this into $f(x)$:
$f(x) = (\sin^{-1} x)^2 + (\frac{\pi}{2} - \sin^{-1} x)^2$.
Let $t = \sin^{-1} x$. Since $x \in [-1, 1]$, $t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
$f(t) = t^2 + (\frac{\pi}{2} - t)^2 = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$.
This is a parabola opening upwards. The vertex is at $t = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$.
Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, the minimum value is $f(\frac{\pi}{4}) = 2(\frac{\pi}{4})^2 - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$.
The maximum value occurs at the endpoints of the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
$f(-\frac{\pi}{2}) = 2(-\frac{\pi}{2})^2 - \pi(-\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$.
$f(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 - \pi(\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}$.
Thus, the greatest value is $\frac{5\pi^2}{4}$ and the least value is $\frac{\pi^2}{8}$.
Sum $= \frac{5\pi^2}{4} + \frac{\pi^2}{8} = \frac{10\pi^2 + \pi^2}{8} = \frac{11\pi^2}{8}$.

(A) We know that $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$.
Let $a = \tan^{-1} x$ and $b = \cot^{-1} x$. Then $a + b = \frac{\pi}{2}$, so $b = \frac{\pi}{2} - a$.
The given equation is $a^2 + b^2 = \frac{5\pi^2}{8}$.
Substituting $b$, we get $a^2 + (\frac{\pi}{2} - a)^2 = \frac{5\pi^2}{8}$.
$a^2 + \frac{\pi^2}{4} - \pi a + a^2 = \frac{5\pi^2}{8}$.
$2a^2 - \pi a + \frac{2\pi^2}{8} - \frac{5\pi^2}{8} = 0$.
$2a^2 - \pi a - \frac{3\pi^2}{8} = 0$.
Multiply by $8$: $16a^2 - 8\pi a - 3\pi^2 = 0$.
Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we get $a = \frac{8\pi \pm \sqrt{64\pi^2 - 4(16)(-3\pi^2)}}{32} = \frac{8\pi \pm \sqrt{64\pi^2 + 192\pi^2}}{32} = \frac{8\pi \pm \sqrt{256\pi^2}}{32} = \frac{8\pi \pm 16\pi}{32}$.
Case $1$: $a = \frac{24\pi}{32} = \frac{3\pi}{4}$. Since the range of $\tan^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, this is invalid.
Case $2$: $a = \frac{-8\pi}{32} = -\frac{\pi}{4}$.
Thus, $\tan^{-1} x = -\frac{\pi}{4}$, which implies $x = \tan(-\frac{\pi}{4}) = -1$.

(A) Let $u = \sin^{-1} x$. Since $0 \le x \le 1$, we have $0 \le u \le \frac{\pi}{2}$.
Then $\cos^{-1} x = \frac{\pi}{2} - u$.
The expression becomes $f(u) = u^3 + (\frac{\pi}{2} - u)^3$ for $u \in [0, \frac{\pi}{2}]$.
$f(u) = u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8}$.
To find the range, find the critical points: $f'(u) = 3\pi u - \frac{3\pi^2}{4} = 0 \implies u = \frac{\pi}{4}$.
At $u = 0$, $f(0) = \frac{\pi^3}{8}$.
At $u = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = (\frac{\pi}{4})^3 + (\frac{\pi}{4})^3 = 2(\frac{\pi^3}{64}) = \frac{\pi^3}{32}$.
At $u = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = (\frac{\pi}{2})^3 + 0 = \frac{\pi^3}{8}$.
Thus, the minimum value is $\frac{\pi^3}{32}$ and the maximum value is $\frac{\pi^3}{8}$.
Given $a\pi^3 = f(u)$, we have $\frac{\pi^3}{32} \le a\pi^3 \le \frac{\pi^3}{8}$, so $\frac{1}{32} \le a \le \frac{1}{8}$.
This implies $a \ge \frac{1}{32}$.

(B) Given $\sin(\cos^{-1} x + \tan^{-1} x) - \cos(\sin^{-1} x + \tan^{-1} x) = \sin(\cot^{-1} 2)$.
Let $\cos^{-1} x = \alpha$, $\sin^{-1} x = \beta$, and $\tan^{-1} x = \gamma$.
We know that $\alpha + \beta = \frac{\pi}{2}$, so $\beta = \frac{\pi}{2} - \alpha$.
The $LHS$ becomes:
$\sin(\alpha + \gamma) - \cos(\beta + \gamma)$
$= \sin(\alpha + \gamma) - \cos(\frac{\pi}{2} - \alpha + \gamma)$
$= \sin(\alpha + \gamma) - \sin(\alpha - \gamma)$
$= 2 \cos \alpha \sin \gamma$.
Since $\cos^{-1} x = \alpha$, we have $\cos \alpha = x$.
Since $\tan^{-1} x = \gamma$, we have $\tan \gamma = x$, which gives $\sin \gamma = \frac{x}{\sqrt{1+x^2}}$.
So, $LHS$ $= 2(x)\left(\frac{x}{\sqrt{1+x^2}}\right) = \frac{2x^2}{\sqrt{1+x^2}}$.
For the $RHS$, let $\cot^{-1} 2 = \delta$. Then $\cot \delta = 2$, which gives $\sin \delta = \frac{1}{\sqrt{1+2^2}} = \frac{1}{\sqrt{5}}$.
Equating $LHS$ and $RHS$:
$\frac{2x^2}{\sqrt{1+x^2}} = \frac{1}{\sqrt{5}}$
Squaring both sides:
$\frac{4x^4}{1+x^2} = \frac{1}{5}$
$20x^4 - x^2 - 1 = 0$
$(5x^2 + 1)(4x^2 - 1) = 0$
Since $x^2 > 0$, we have $4x^2 - 1 = 0 \implies x^2 = \frac{1}{4}$.
Given $x > 0$, we get $x = \frac{1}{2}$.

(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.
Let $A = \sin^{-1} x$ and $B = \cos^{-1} x$. Then $A + B = \frac{\pi}{2}$, which implies $B = \frac{\pi}{2} - A$.
The given equation is $2A - 3B = 4$.
Substitute $B = \frac{\pi}{2} - A$ into the equation: $2A - 3(\frac{\pi}{2} - A) = 4$.
$2A - \frac{3\pi}{2} + 3A = 4 \implies 5A = 4 + \frac{3\pi}{2} = \frac{8 + 3\pi}{2} \implies A = \frac{8 + 3\pi}{10}$.
Now find $B$: $B = \frac{\pi}{2} - \frac{8 + 3\pi}{10} = \frac{5\pi - 8 - 3\pi}{10} = \frac{2\pi - 8}{10} = \frac{\pi - 4}{5}$.
We need to find $2A + 3B$: $2(\frac{8 + 3\pi}{10}) + 3(\frac{2\pi - 8}{10}) = \frac{16 + 6\pi + 6\pi - 24}{10} = \frac{12\pi - 8}{10} = \frac{6\pi - 4}{5}$.

(C) Let $\cos^{-1} x = \theta$, then $\cos \theta = x$. Thus, $\cot(\cos^{-1} x) = \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{\sqrt{1 - x^2}}$.
Let $\tan^{-1} \frac{a}{\sqrt{b^2 - a^2}} = \phi$, then $\tan \phi = \frac{a}{\sqrt{b^2 - a^2}}$.
Using the identity $\sec^2 \phi = 1 + \tan^2 \phi$, we get $\sec^2 \phi = 1 + \frac{a^2}{b^2 - a^2} = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2}$.
Therefore, $\sec \phi = \frac{b}{\sqrt{b^2 - a^2}}$.
Equating the two sides: $\frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}$.
Squaring both sides: $\frac{x^2}{1 - x^2} = \frac{b^2}{b^2 - a^2}$.
$x^2(b^2 - a^2) = b^2(1 - x^2) = b^2 - b^2 x^2$.
$x^2(b^2 - a^2 + b^2) = b^2 \implies x^2(2b^2 - a^2) = b^2$.
$x^2 = \frac{b^2}{2b^2 - a^2} \implies x = \frac{b}{\sqrt{2b^2 - a^2}}$.

(B) Given: $\tan^{-1} \left[ \frac{\sqrt{5} - 2\sqrt{6}}{1 + \sqrt{6}} \right] = \frac{\pi}{3} - \tan^{-1}(k)$.
Note that $\sqrt{5} - 2\sqrt{6} = \sqrt{3} - \sqrt{2}$ is incorrect; let us simplify the expression inside $\tan^{-1}$.
Actually, $\frac{\sqrt{3} - \sqrt{2}}{1 + \sqrt{6}} = \tan(\tan^{-1}(\sqrt{3}) - \tan^{-1}(\sqrt{2})) = \tan(\frac{\pi}{3} - \tan^{-1}(\sqrt{2}))$.
Comparing this with the given equation, we have $\tan^{-1}(\sqrt{2}) = \tan^{-1}(k)$, so $k = \sqrt{2}$.
We need to find $\sec^{-1}(\sqrt{2})$.
Since $\sec(\frac{\pi}{4}) = \sqrt{2}$, it follows that $\sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$.
Thus, the correct option is $B$.

(D) Given $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} \right)$.
Using trigonometric identities $1 + \sin \theta = \left( \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \right)^2$ and $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}$, we have:
$y = \cot^{-1} \left( \frac{(\cos \frac{5x}{2} + \sin \frac{5x}{2})^2}{(\cos \frac{5x}{2} - \sin \frac{5x}{2})(\cos \frac{5x}{2} + \sin \frac{5x}{2})} \right)$
$y = \cot^{-1} \left( \frac{\cos \frac{5x}{2} + \sin \frac{5x}{2}}{\cos \frac{5x}{2} - \sin \frac{5x}{2}} \right)$
Divide numerator and denominator by $\cos \frac{5x}{2}$:
$y = \cot^{-1} \left( \frac{1 + \tan \frac{5x}{2}}{1 - \tan \frac{5x}{2}} \right) = \cot^{-1} \left( \tan \left( \frac{\pi}{4} + \frac{5x}{2} \right) \right)$
Since $\cot^{-1}(\tan \theta) = \cot^{-1}(\cot(\frac{\pi}{2} - \theta))$, we get:
$y = \frac{\pi}{2} - (\frac{\pi}{4} + \frac{5x}{2}) = \frac{\pi}{4} - \frac{5x}{2}$
Therefore, $\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{4} - \frac{5x}{2}) = -\frac{5}{2}$.

(D) Let $x = \cos \theta$, then $\theta = \cos^{-1} x$.
$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} = \sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}} = \tan(\theta/2)$.
Now, $\cot^{-1}(\tan(\theta/2)) = \cot^{-1}(\cot(\frac{\pi}{2} - \frac{\theta}{2})) = \frac{\pi}{2} - \frac{\theta}{2}$.
Substituting this into the expression for $y$:
$y = \cos^2(\frac{\pi}{2} - \frac{\theta}{2}) = \sin^2(\theta/2)$.
Using the identity $\sin^2(\theta/2) = \frac{1-\cos \theta}{2}$:
$y = \frac{1-x}{2} = \frac{1}{2} - \frac{x}{2}$.
Therefore, $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} - \frac{x}{2}) = -\frac{1}{2}$.

(A) The general term of the series is $T_k = \tan^{-1} \left( \frac{1}{1 + k(k + 1)} \right)$.
We can rewrite the argument as $\frac{(k + 1) - k}{1 + k(k + 1)}$.
Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x - y}{1 + xy} \right)$, we get $T_k = \tan^{-1}(k + 1) - \tan^{-1}(k)$.
The sum $S_n = \sum_{k=1}^{n} (\tan^{-1}(k + 1) - \tan^{-1}(k))$.
This is a telescoping series: $S_n = (\tan^{-1} 2 - \tan^{-1} 1) + (\tan^{-1} 3 - \tan^{-1} 2) + \dots + (\tan^{-1}(n + 1) - \tan^{-1} n)$.
All intermediate terms cancel out, leaving $S_n = \tan^{-1}(n + 1) - \tan^{-1}(1)$.
Using the identity again, $S_n = \tan^{-1} \left( \frac{(n + 1) - 1}{1 + (n + 1)(1)} \right) = \tan^{-1} \left( \frac{n}{n + 2} \right)$.

(A) Given $\sin^{-1} x + \sin^{-1} y = \pi/2$.
We know that $\sin^{-1} y + \cos^{-1} y = \pi/2$, so $\sin^{-1} y = \pi/2 - \cos^{-1} y$.
Substituting this into the given equation: $\sin^{-1} x = \pi/2 - \sin^{-1} y = \cos^{-1} y$.
Taking $\sin$ on both sides: $x = \sin(\cos^{-1} y)$.
Using the identity $\sin(\cos^{-1} y) = \sqrt{1 - y^2}$, we get $x = \sqrt{1 - y^2}$.
Squaring both sides, we get $x^2 = 1 - y^2$.