Properties of ITF Questions in English

(A) Given that $x, y, z$ are in $A.P.$
$\therefore 2y = x + z$ ... $(i)$
Also,$\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in $A.P.$
$\therefore 2 \tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} z$
Using the formula $\tan ^{-1} a + \tan ^{-1} b = \tan ^{-1} \left( \frac{a+b}{1-ab} \right)$:
$\tan ^{-1} \left( \frac{2y}{1-y^2} \right) = \tan ^{-1} \left( \frac{x+z}{1-xz} \right)$
$\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$
Substituting $x+z = 2y$ from $(i)$:
$\frac{2y}{1-y^2} = \frac{2y}{1-xz}$
This implies either $2y = 0$ (which leads to $x=y=z=0$) or $1-y^2 = 1-xz$,which gives $y^2 = xz$.
Since $x, y, z$ are in both $A.P.$ and $G.P.$,we must have $x=y=z$.

(A) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$.
Using the identity $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^2})$,we have:
$\cos ^{-1} \left( \frac{xy}{3} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{9}} \right) = \alpha$
$\frac{xy}{3} + \frac{\sqrt{1-x^2} \sqrt{9-y^2}}{3} = \cos \alpha$
$xy + \sqrt{1-x^2} \sqrt{9-y^2} = 3 \cos \alpha$
$xy - 3 \cos \alpha = -\sqrt{1-x^2} \sqrt{9-y^2}$
Squaring both sides:
$(xy - 3 \cos \alpha)^2 = (1-x^2)(9-y^2)$
$x^2y^2 - 6xy \cos \alpha + 9 \cos ^2 \alpha = 9 - y^2 - 9x^2 + x^2y^2$
$9x^2 - 6xy \cos \alpha + y^2 = 9 - 9 \cos ^2 \alpha$
$9x^2 - 6xy \cos \alpha + y^2 = 9(1 - \cos ^2 \alpha) = 9 \sin ^2 \alpha$.

(A) Let $\theta = \frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}$.
Then $2\theta = \cos ^{-1} \frac{\sqrt{5}}{3}$,which implies $\cos 2\theta = \frac{\sqrt{5}}{3}$.
We need to find $\tan \theta$.
Using the formula $\tan \theta = \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}$:
$\tan \theta = \sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}$.
Rationalizing the denominator:
$\tan \theta = \sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}} = \sqrt{\frac{(3-\sqrt{5})^2}{9-5}} = \sqrt{\frac{(3-\sqrt{5})^2}{4}} = \frac{3-\sqrt{5}}{2}$.

(A) Given equation: $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$
Let $\cot ^{-1}(x+1) = \theta$,then $\cot \theta = x+1$. Using the identity $\sin \theta = \frac{1}{\sqrt{1+\cot^2 \theta}}$,we get $\sin \theta = \frac{1}{\sqrt{1+(x+1)^2}} = \frac{1}{\sqrt{x^2+2x+2}}$.
Let $\tan ^{-1} x = \phi$,then $\tan \phi = x$. Using the identity $\cos \phi = \frac{1}{\sqrt{1+\tan^2 \phi}}$,we get $\cos \phi = \frac{1}{\sqrt{1+x^2}}$.
Equating both sides: $\frac{1}{\sqrt{x^2+2x+2}} = \frac{1}{\sqrt{x^2+1}}$.
Squaring both sides: $x^2+1 = x^2+2x+2$.
$2x = -1 \Rightarrow x = -\frac{1}{2}$.

(D) Given $\tan \alpha = \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$.
Let $x^2 = \cos \theta$. Then $\theta = \cos^{-1}(x^2)$.
Substituting $x^2 = \cos \theta$ into the expression:
$\tan \alpha = \frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}$
Using half-angle identities $1+\cos \theta = 2\cos^2(\theta/2)$ and $1-\cos \theta = 2\sin^2(\theta/2)$:
$\tan \alpha = \frac{\sqrt{2}\cos(\theta/2) - \sqrt{2}\sin(\theta/2)}{\sqrt{2}\cos(\theta/2) + \sqrt{2}\sin(\theta/2)}$
Dividing numerator and denominator by $\sqrt{2}\cos(\theta/2)$:
$\tan \alpha = \frac{1 - \tan(\theta/2)}{1 + \tan(\theta/2)} = \tan(\frac{\pi}{4} - \frac{\theta}{2})$
Thus,$\alpha = \frac{\pi}{4} - \frac{\theta}{2}$,which implies $2\alpha = \frac{\pi}{2} - \theta$.
Therefore,$\sin 2\alpha = \sin(\frac{\pi}{2} - \theta) = \cos \theta$.
Since $x^2 = \cos \theta$,we have $\sin 2\alpha = x^2$.

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

(A) The $n^{th}$ term of the series is $T_n = \tan^{-1}\left(\frac{2^{n-1}}{1+2^{2n-1}}\right)$.
We can rewrite the argument as $\frac{2^n - 2^{n-1}}{1 + 2^n \cdot 2^{n-1}}$.
Using the formula $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$,we get $T_n = \tan^{-1}(2^n) - \tan^{-1}(2^{n-1})$.
The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (\tan^{-1}(2^k) - \tan^{-1}(2^{k-1}))$.
This is a telescoping series: $S_n = (\tan^{-1}(2^1) - \tan^{-1}(2^0)) + (\tan^{-1}(2^2) - \tan^{-1}(2^1)) + \dots + (\tan^{-1}(2^n) - \tan^{-1}(2^{n-1}))$.
$S_n = \tan^{-1}(2^n) - \tan^{-1}(1) = \tan^{-1}(2^n) - \frac{\pi}{4}$.
As $n \to \infty$,$S_n = \tan^{-1}(\infty) - \frac{\pi}{4} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

(C) Let $E = \sin \left[3 \sin ^{-1}(0.4)\right]$.
Put $\sin ^{-1}(0.4) = \theta$,which implies $\sin \theta = 0.4$.
Using the identity $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$:
$E = 3(0.4) - 4(0.4)^3$
$E = 1.2 - 4(0.064)$
$E = 1.2 - 0.256$
$E = 0.944$

(B) Given the equation: $\cos \left(\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x\right)=1$
Taking $\cos ^{-1}$ on both sides:
$\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x = \cos ^{-1}(1)$
We know that $\cos ^{-1}(1) = 0$ and $\cos ^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}$.
Substituting these values:
$\frac{\pi}{6} + \sin ^{-1} x = 0$
$\sin ^{-1} x = -\frac{\pi}{6}$
$x = \sin \left(-\frac{\pi}{6}\right)$
Since $\sin(-\theta) = -\sin(\theta)$,we have:
$x = -\sin \left(\frac{\pi}{6}\right) = -\frac{1}{2}$
Thus,the value of $x$ is $-\frac{1}{2}$.

(D) We need to evaluate the sum $S = \sum_{i=1}^3 \cot ^{-1} i = \cot ^{-1} 1 + \cot ^{-1} 2 + \cot ^{-1} 3$.
Using the identity $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$,we have $S = \tan ^{-1} 1 + \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{3}$.
Since $\tan ^{-1} 1 = \frac{\pi}{4}$,we calculate $\tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{3} = \tan ^{-1} \left( \frac{1/2 + 1/3}{1 - (1/2)(1/3)} \right) = \tan ^{-1} \left( \frac{5/6}{5/6} \right) = \tan ^{-1} 1 = \frac{\pi}{4}$.
Thus,$S = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.
Now,we substitute this into the expression: $\cos ^{-1} \{ \cot (S) \} = \cos ^{-1} \{ \cot (\frac{\pi}{2}) \}$.
Since $\cot (\frac{\pi}{2}) = 0$,the expression becomes $\cos ^{-1} (0)$.
We know that $\cos ^{-1} (0) = \frac{\pi}{2}$.
Therefore,the correct option is $D$.

(B) Given the equation: $\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=0$.
We know that $\cos \theta = 0$ implies $\theta = \frac{\pi}{2}$ (considering the principal value branch).
Therefore,$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$.
Rearranging the terms,we get $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} \frac{1}{5}$.
Using the identity $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$,we have $\cos ^{-1} y = \frac{\pi}{2} - \sin ^{-1} y$.
Thus,$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$.
Comparing both sides,we get $x = \frac{1}{5}$.

(C) We know that $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$ holds true when $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.
Let $x = \sin \theta$,where $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Then $2 \sin^{-1}(\sin \theta) = \sin^{-1}(2 \sin \theta \sqrt{1-\sin^2 \theta}) = \sin^{-1}(2 \sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta)$.
For $2\theta = 2 \sin^{-1} x$ to be valid,$2\theta$ must lie in the principal value branch of $\sin^{-1}$,which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
So,$-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2} \implies -\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$.
Taking sine on all sides,$\sin(-\frac{\pi}{4}) \le \sin \theta \le \sin(\frac{\pi}{4})$,which gives $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.
Thus,$x \in [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$.

(C) Given that $\sin ^{-1} a = \alpha + \beta$ and $\sin ^{-1} b = \alpha - \beta$.
This implies $a = \sin(\alpha + \beta)$ and $b = \sin(\alpha - \beta)$.
Using the product-to-sum formula for sine:
$a b = \sin(\alpha + \beta) \sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta$.
We need to find $\sin^2 \alpha + \cos^2 \beta$.
Since $\cos^2 \beta = 1 - \sin^2 \beta$,we substitute this into the expression:
$\sin^2 \alpha + \cos^2 \beta = \sin^2 \alpha + (1 - \sin^2 \beta) = 1 + (\sin^2 \alpha - \sin^2 \beta)$.
Substituting the value of $ab$ from the product formula:
$1 + (\sin^2 \alpha - \sin^2 \beta) = 1 + a b$.
Therefore,the correct option is $C$.

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$ and $\beta = \sin^{-1}\left(\frac{8}{17}\right)$.
Then $\sin \alpha = \frac{3}{5}$ and $\sin \beta = \frac{8}{17}$.
Using the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$,we get $\cos \alpha = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ and $\cos \beta = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.
We use the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
$\cos(\alpha - \beta) = \left(\frac{4}{5}\right)\left(\frac{15}{17}\right) + \left(\frac{3}{5}\right)\left(\frac{8}{17}\right) = \frac{60}{85} + \frac{24}{85} = \frac{84}{85}$.
Therefore,$\alpha - \beta = \cos^{-1}\left(\frac{84}{85}\right)$.

(D) Let the expression be $E = \tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})$.
Step $1$: Simplify $\tan ^2(\sec ^{-1} 3)$.
Let $\sec ^{-1} 3 = \theta_1$,so $\sec \theta_1 = 3$. Then $\tan ^2 \theta_1 = \sec ^2 \theta_1 - 1 = 3^2 - 1 = 9 - 1 = 8$.
Step $2$: Simplify $\operatorname{cosec}^2(\cot ^{-1} 2)$.
Let $\cot ^{-1} 2 = \theta_2$,so $\cot \theta_2 = 2$. Then $\operatorname{cosec}^2 \theta_2 = 1 + \cot ^2 \theta_2 = 1 + 2^2 = 1 + 4 = 5$.
Step $3$: Simplify $\cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})$.
We know that $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Thus,$\cos ^2(\frac{\pi}{2}) = 0^2 = 0$.
Step $4$: Calculate the total sum.
$E = 8 + 5 + 0 = 13$.
Therefore,the correct option is $D$.

(C) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Let $\sin^{-1} x = \theta$,then $x = \sin \theta$. Thus,$\cos(\sin^{-1} x) = \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}$.
So,$\sin^{-1}(\cos(\sin^{-1} x)) = \sin^{-1}(\sqrt{1 - x^2})$.
Similarly,let $\cos^{-1} x = \phi$,then $x = \cos \phi$. Thus,$\sin(\cos^{-1} x) = \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - x^2}$.
So,$\cos^{-1}(\sin(\cos^{-1} x)) = \cos^{-1}(\sqrt{1 - x^2})$.
Now,the expression becomes $\sin^{-1}(\sqrt{1 - x^2}) + \cos^{-1}(\sqrt{1 - x^2})$.
Using the identity $\sin^{-1} A + \cos^{-1} A = \frac{\pi}{2}$ where $A = \sqrt{1 - x^2}$,we get the result $\frac{\pi}{2}$.

(C) Let $x = \tan \theta$,where $\theta = \tan^{-1} x$.
Then,$\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta$.
The expression becomes $\cot^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$.
$= \cot^{-1}\left(\frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}}\right) = \cot^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right)$.
Using half-angle formulas,$1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$.
$= \cot^{-1}\left(\frac{2 \sin^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) = \cot^{-1}\left(\tan \frac{\theta}{2}\right)$.
Since $\cot^{-1}(\tan \alpha) = \frac{\pi}{2} - \tan^{-1}(\tan \alpha) = \frac{\pi}{2} - \alpha$,we have:
$= \frac{\pi}{2} - \frac{\theta}{2} = \frac{\pi}{2} - \frac{1}{2} \tan^{-1} x$.

(D) We know that $\tan ^{-1} x + \tan ^{-1} \frac{1}{x} = \frac{\pi}{2}$ for $x > 0$.
Given expression is $\cos \left(2 \left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} 5\right)\right)$.
Since $5 > 0$,we have $\tan ^{-1} \frac{1}{5} + \tan ^{-1} 5 = \frac{\pi}{2}$.
Substituting this into the expression:
$\cos \left(2 \times \frac{\pi}{2}\right) = \cos(\pi)$.
We know that $\cos(\pi) = -1$.
Therefore,the correct option is $D$.

(D) We know that $\cot x = \tan(\frac{\pi}{2} - x)$ and $\tan x = \cot(\frac{\pi}{2} - x)$.
Substituting these into the expression,we get:
$\tan^{-1}(\tan(\frac{\pi}{2} - x)) + \cot^{-1}(\cot(\frac{\pi}{2} - x))$
$= (\frac{\pi}{2} - x) + (\frac{\pi}{2} - x)$
$= \pi - 2x$.
Therefore,the correct option is $D$.

(A) Given the equation: $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$
We know that $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$ for $y \in [-1, 1]$.
Thus,$\sin ^{-1} \frac{x}{5} = \frac{\pi}{2} - \sin ^{-1} \frac{4}{5}$.
Using the identity,we get $\sin ^{-1} \frac{x}{5} = \cos ^{-1} \frac{4}{5}$.
Let $\cos ^{-1} \frac{4}{5} = \theta$,then $\cos \theta = \frac{4}{5}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.
Therefore,$\sin ^{-1} \frac{x}{5} = \sin ^{-1} \frac{3}{5}$.
Comparing both sides,we get $\frac{x}{5} = \frac{3}{5}$,which implies $x = 3$.

(C) We know that $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
Also,we use the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$.
Therefore,$\cot^{-1}(-\sqrt{3}) = \pi - \cot^{-1}(\sqrt{3})$.
Since $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$,we have $\cot^{-1}(-\sqrt{3}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Now,substituting these values into the expression:
$\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) = \frac{\pi}{3} - \frac{5\pi}{6}$.
Finding a common denominator:
$\frac{2\pi}{6} - \frac{5\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$.
Thus,the correct option is $C$.

(A) Let $x = \sec \theta$. Since $x > 1$,we have $0 < \theta < \frac{\pi}{2}$.
Then,$\sqrt{x^2-1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta} = \tan \theta$.
Substituting this into the expression,we get $\cot ^{-1}\left(\frac{1}{\tan \theta}\right) = \cot ^{-1}(\cot \theta)$.
Since $0 < \theta < \frac{\pi}{2}$,$\cot ^{-1}(\cot \theta) = \theta$.
Substituting back $\theta = \sec ^{-1} x$,we get the result $\sec ^{-1} x$.

(A) We know the identity for inverse trigonometric functions: $\sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}$ for all $|x| \geq 1$.
Substituting this into the given expression:
$\cos \left[\sec^{-1} x + \operatorname{cosec}^{-1} x\right] = \cos \left(\frac{\pi}{2}\right)$.
Since $\cos \left(\frac{\pi}{2}\right) = 0$,the value of the expression is $0$.

(C) We know that $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$ for $x > 0$.
Given expression is $\cot ^{-1}(-\sqrt{3}) - \tan ^{-1}(\sqrt{3})$.
Using the property,$\cot ^{-1}(-\sqrt{3}) = \pi - \cot ^{-1}(\sqrt{3})$.
Since $\cot ^{-1}(\sqrt{3}) = \frac{\pi}{6}$ and $\tan ^{-1}(\sqrt{3}) = \frac{\pi}{3}$,
Substituting these values:
$\cot ^{-1}(-\sqrt{3}) - \tan ^{-1}(\sqrt{3}) = (\pi - \frac{\pi}{6}) - \frac{\pi}{3}$.
$= \frac{5\pi}{6} - \frac{\pi}{3} = \frac{5\pi - 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.
Thus,the correct option is $C$.

(A) We know the standard identity for inverse trigonometric functions: $\sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}$ for $|x| \geq 1$.
Similarly,for $|x| \geq 1$,we have $\cos^{-1}(x^{-1}) + \sin^{-1}(x^{-1}) = \frac{\pi}{2}$.
Substituting these identities into the given expression:
$\sec^{-1} x + \operatorname{cosec}^{-1} x + \cos^{-1}(x^{-1}) + \sin^{-1}(x^{-1}) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
Thus,the correct option is $A$.

(D) We need to evaluate the sum $S = \sum_{i=0}^2 \cot^{-1}(-(i+1))$.
Expanding the sum,we get:
$S = \cot^{-1}(-1) + \cot^{-1}(-2) + \cot^{-1}(-3)$.
Using the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$:
$S = (\pi - \cot^{-1}(1)) + (\pi - \cot^{-1}(2)) + (\pi - \cot^{-1}(3))$
$S = 3\pi - (\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3))$.
We know $\cot^{-1}(1) = \frac{\pi}{4}$.
For $\cot^{-1}(2) + \cot^{-1}(3)$,we use $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})$ where $xy < 1$:
$\cot^{-1}(2) + \cot^{-1}(3) = \tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{1}{3}) = \tan^{-1}(\frac{1/2 + 1/3}{1 - 1/6}) = \tan^{-1}(\frac{5/6}{5/6}) = \tan^{-1}(1) = \frac{\pi}{4}$.
Thus,$S = 3\pi - (\frac{\pi}{4} + \frac{\pi}{4}) = 3\pi - \frac{\pi}{2} = \frac{5\pi}{2}$.

(A) We know that for any $x \in [-1, 1]$,the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ holds true.
Given the expression $\cos \left(\cos ^{-1}\left(-\frac{1}{4}\right)+\sin ^{-1}\left(-\frac{1}{4}\right)\right)$.
Let $x = -\frac{1}{4}$. Since $x \in [-1, 1]$,we can apply the identity.
Therefore,$\cos^{-1}\left(-\frac{1}{4}\right) + \sin^{-1}\left(-\frac{1}{4}\right) = \frac{\pi}{2}$.
Substituting this into the original expression,we get $\cos \left(\frac{\pi}{2}\right)$.
Since $\cos \left(\frac{\pi}{2}\right) = 0$,the final answer is $0$.

(D) We know that $\cot ^{-1}(x) = \tan ^{-1}\left(\frac{1}{x}\right)$ for $x > 0$.
Given expression is $\cot ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{1}{3}\right)$.
This can be written as $\tan ^{-1}(2) + \tan ^{-1}(3)$.
Since $xy = 2 \times 3 = 6 > 1$,we use the formula $\tan ^{-1}(x) + \tan ^{-1}(y) = \pi + \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$.
Substituting the values,we get $\pi + \tan ^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan ^{-1}\left(\frac{5}{-5}\right)$.
This simplifies to $\pi + \tan ^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.
Thus,the correct option is $D$.

(A) We are given the equation $3 \cos ^{-1} x + \sin ^{-1} x = \pi$.
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$.
Substituting this into the given equation:
$3 \cos ^{-1} x + (\frac{\pi}{2} - \cos ^{-1} x) = \pi$
$2 \cos ^{-1} x + \frac{\pi}{2} = \pi$
$2 \cos ^{-1} x = \pi - \frac{\pi}{2}$
$2 \cos ^{-1} x = \frac{\pi}{2}$
$\cos ^{-1} x = \frac{\pi}{4}$
$x = \cos(\frac{\pi}{4})$
$x = \frac{1}{\sqrt{2}}$
Therefore,the correct option is $A$.

(B) We know the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Here,let $x = -\frac{1}{7}$.
Since $-\frac{1}{7} \in [-1, 1]$,we can apply the identity.
Therefore,$\cos^{-1}\left(-\frac{1}{7}\right) + \sin^{-1}\left(-\frac{1}{7}\right) = \frac{\pi}{2}$.
Substituting this into the original expression,we get $\sin\left(\frac{\pi}{2}\right)$.
Since $\sin\left(\frac{\pi}{2}\right) = 1$,the final answer is $1$.

(C) Given that $\tan ^{-1}(1-x), \tan ^{-1} x$,and $\tan ^{-1}(1+x)$ are in $A$.$P$.
Therefore,$2 \tan ^{-1} x = \tan ^{-1}(1-x) + \tan ^{-1}(1+x)$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$2 \tan ^{-1} x = \tan ^{-1} \left( \frac{(1-x) + (1+x)}{1 - (1-x)(1+x)} \right)$.
$2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2}{1 - (1-x^2)} \right)$.
$2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2}{x^2} \right)$.
Using the formula $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$,we have:
$\frac{2x}{1-x^2} = \frac{2}{x^2}$.
$x^3 = 1 - x^2$.
Thus,$x^3 = 1 - x^2$.

(D) Given the equation: $\sec ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1} \left(\frac{4}{5}\right)=\frac{\pi}{2}$.
We know the identity $\sin ^{-1}(y) + \cos ^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$.
Thus,$\frac{\pi}{2} - \sin ^{-1} \left(\frac{4}{5}\right) = \cos ^{-1} \left(\frac{4}{5}\right)$.
Substituting this into the original equation,we get: $\sec ^{-1}\left(\frac{5}{x}\right) = \cos ^{-1} \left(\frac{4}{5}\right)$.
Since $\sec ^{-1}(z) = \cos ^{-1} \left(\frac{1}{z}\right)$,we have $\cos ^{-1} \left(\frac{x}{5}\right) = \cos ^{-1} \left(\frac{4}{5}\right)$.
Comparing the arguments,we get $\frac{x}{5} = \frac{4}{5}$,which implies $x = 4$.

(D) We use the trigonometric identities: $\sec ^2 \theta = 1 + \tan ^2 \theta$ and $\operatorname{cosec}^2 \theta = 1 + \cot ^2 \theta$.
Let $\alpha = \tan ^{-1} 2$,then $\tan \alpha = 2$.
Let $\beta = \cot ^{-1} 3$,then $\cot \beta = 3$.
The expression becomes:
$\sec ^2(\tan ^{-1} 2) + \operatorname{cosec}^2(\cot ^{-1} 3) = (1 + \tan ^2(\tan ^{-1} 2)) + (1 + \cot ^2(\cot ^{-1} 3))$
$= (1 + 2^2) + (1 + 3^2)$
$= (1 + 4) + (1 + 9)$
$= 5 + 10 = 15$.
Thus,the correct option is $D$.

(A) We are given the equation: $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$.
We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,which implies $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$.
Substituting this into the given equation:
$4 \cos^{-1} x + (\frac{\pi}{2} - \cos^{-1} x) = \frac{\pi}{2}$.
Simplifying the expression:
$3 \cos^{-1} x + \frac{\pi}{2} = \frac{\pi}{2}$.
Subtracting $\frac{\pi}{2}$ from both sides:
$3 \cos^{-1} x = 0$.
Dividing by $3$:
$\cos^{-1} x = 0$.
Taking the cosine of both sides:
$x = \cos(0) = 1$.

(A) Let $x = \cos^{-1} \frac{1}{3} + \cos^{-1} \frac{1}{5}$ and $y = \sin^{-1} \frac{1}{3} + \sin^{-1} \frac{1}{5}$.
We know that $\cos^{-1} z = \frac{\pi}{2} - \sin^{-1} z$.
Thus,$x = (\frac{\pi}{2} - \sin^{-1} \frac{1}{3}) + (\frac{\pi}{2} - \sin^{-1} \frac{1}{5}) = \pi - (\sin^{-1} \frac{1}{3} + \sin^{-1} \frac{1}{5}) = \pi - y$.
Therefore,$\cos(x) = \cos(\pi - y) = -\cos(y)$.
This implies $\cos(x) + \cos(y) = -\cos(y) + \cos(y) = 0$.

(C) We know that $\cos \theta = \sin \left(\frac{\pi}{2} - \theta\right)$ and $\sin \theta = \cos \left(\frac{\pi}{2} - \theta\right)$.
Substituting these into the expression:
$\sin ^{-1}\left(\sin \left(\frac{\pi}{2} - \frac{\pi}{13}\right)\right) + \cos ^{-1}\left(\cos \left(\frac{\pi}{2} - \frac{\pi}{13}\right)\right)$
Since $\frac{\pi}{2} - \frac{\pi}{13} = \frac{11\pi}{26}$,which lies in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}$ and $[0, \pi]$ for $\cos^{-1}$,we have:
$= \left(\frac{\pi}{2} - \frac{\pi}{13}\right) + \left(\frac{\pi}{2} - \frac{\pi}{13}\right)$
$= \pi - \frac{2\pi}{13}$
$= \frac{13\pi - 2\pi}{13} = \frac{11\pi}{13}$.

(A) Let $y = \cos^{-1}(\sin x)$.
We know that $\sin x = \cos(\frac{\pi}{2} - x)$.
Substituting this into the expression,we get $y = \cos^{-1}(\cos(\frac{\pi}{2} - x))$.
Since $\cos^{-1}(\cos \theta) = \theta$ for $\theta \in [0, \pi]$,we have $y = \frac{\pi}{2} - x$.
Now,differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{2} - x) = 0 - 1 = -1$.
Thus,the derivative is $-1$.

(D) Given the equation: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1} x$.
We know that $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \tan^{-1}(1) - \tan^{-1}(x)$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$,the equation becomes: $\frac{\pi}{4} - \tan^{-1} x = \frac{1}{2} \tan^{-1} x$.
Adding $\tan^{-1} x$ to both sides: $\frac{\pi}{4} = \frac{1}{2} \tan^{-1} x + \tan^{-1} x$.
$\frac{\pi}{4} = \frac{3}{2} \tan^{-1} x$.
$\tan^{-1} x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.
Therefore,$x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

(B) Let $\alpha = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then $\sin \alpha = \frac{2\sqrt{2}}{3}$.
Since $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{8}{9} = \frac{1}{9}$,we have $\cos \alpha = \frac{1}{3}$.
Thus,$\alpha = \cos^{-1}\left(\frac{1}{3}\right)$.
Substituting this into the expression,we get $\cos^{-1}\left(\frac{1}{3}\right) + \sin^{-1}\left(\frac{1}{3}\right)$.
Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the value is $\frac{\pi}{2}$.

(C) Let the expression be $E = \tan ^{-1}\left[\frac{1}{\sqrt{3}} \sin \frac{5 \pi}{2}\right] + \sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]$.
First,evaluate $\sin \frac{5 \pi}{2} = \sin \left(2 \pi + \frac{\pi}{2}\right) = \sin \frac{\pi}{2} = 1$.
Next,evaluate $\sin ^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{3}$.
Now substitute these values into the expression:
$E = \tan ^{-1}\left[\frac{1}{\sqrt{3}} \times 1\right] + \sin ^{-1}\left[\cos \frac{\pi}{3}\right]$.
Since $\cos \frac{\pi}{3} = \frac{1}{2}$,we have:
$E = \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) + \sin ^{-1}\left(\frac{1}{2}\right)$.
We know that $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ and $\sin ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.
Therefore,$E = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2 \pi}{6} = \frac{\pi}{3}$.

(C) Let $\cos ^{-1} x = y$,then $x = \cos y$. Since the range of $\cos ^{-1} x$ is $[0, \pi]$,we have $0 \leq y \leq \pi$.
The given equation is $2y = \sin ^{-1} (2 \cos y \sin y) = \sin ^{-1} (\sin 2y)$.
For $\sin ^{-1} (\sin 2y) = 2y$,we must have $-\frac{\pi}{2} \leq 2y \leq \frac{\pi}{2}$,which implies $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$.
Combining $0 \leq y \leq \pi$ and $-\frac{\pi}{4} \leq y \leq \frac{\pi}{4}$,we get $0 \leq y \leq \frac{\pi}{4}$.
Since $y = \cos ^{-1} x$,we have $0 \leq \cos ^{-1} x \leq \frac{\pi}{4}$.
Taking cosine on all sides (noting that cosine is a decreasing function in $[0, \pi]$),we get $\cos(0) \geq x \geq \cos(\frac{\pi}{4})$.
Thus,$1 \geq x \geq \frac{1}{\sqrt{2}}$,or $\frac{1}{\sqrt{2}} \leq x \leq 1$.

(D) Given equation: $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$
We know that $\cot ^{-1} x = \tan ^{-1} (\frac{1}{x})$ for $x > 0$.
Substituting this,we get: $\tan ^{-1} x + 2 \tan ^{-1} (\frac{1}{x}) = \frac{2 \pi}{3}$
Using the identity $2 \tan ^{-1} y = \tan ^{-1} (\frac{2y}{1-y^2})$,we have:
$\tan ^{-1} x + \tan ^{-1} (\frac{2/x}{1-1/x^2}) = \frac{2 \pi}{3}$
$\tan ^{-1} x + \tan ^{-1} (\frac{2x}{x^2-1}) = \frac{2 \pi}{3}$
Applying $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} (\frac{A+B}{1-AB})$:
$\tan ^{-1} (\frac{x + \frac{2x}{x^2-1}}{1 - x(\frac{2x}{x^2-1})}) = \frac{2 \pi}{3}$
$\frac{\frac{x^3-x+2x}{x^2-1}}{\frac{x^2-1-2x^2}{x^2-1}} = \tan (\frac{2 \pi}{3})$
$\frac{x^3+x}{-x^2-1} = -\sqrt{3}$
$\frac{x(x^2+1)}{-(x^2+1)} = -\sqrt{3}$
$-x = -\sqrt{3} \implies x = \sqrt{3}$

(D) Let $x = \cos \theta$,where $\theta = \cos^{-1} x$. Since $x \in [-1, 1]$,$\theta \in [0, \pi]$.
Substituting $x = \cos \theta$ into the right-hand side:
$\sin^{-1} (2 \cos \theta \sqrt{1 - \cos^2 \theta}) = \sin^{-1} (2 \cos \theta \sin \theta) = \sin^{-1} (\sin 2 \theta)$.
For this to equal $2 \cos^{-1} x = 2 \theta$,we require $\sin^{-1} (\sin 2 \theta) = 2 \theta$.
This holds when $2 \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,which implies $\theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$.
Given $\theta \in [0, \pi]$,the intersection is $\theta \in [0, \frac{\pi}{4}]$.
Since $\theta = \cos^{-1} x$,we have $0 \leq \cos^{-1} x \leq \frac{\pi}{4}$.
Taking the cosine of all sides (noting that $\cos$ is a decreasing function in this interval):
$\cos(0) \geq x \geq \cos(\frac{\pi}{4})$,
which gives $1 \geq x \geq \frac{1}{\sqrt{2}}$ or $\frac{1}{\sqrt{2}} \leq x \leq 1$.

(C) Let $y = \cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$.
Rationalizing the expression inside the bracket:
$\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(1-\sin x)-(1+\sin x)} = \frac{1-\sin x + 1+\sin x + 2\sqrt{(1-\sin x)(1+\sin x)}}{-2\sin x} = \frac{2 + 2\sqrt{1-\sin^2 x}}{-2\sin x} = \frac{2 + 2\cos x}{-2\sin x} = -\frac{1+\cos x}{\sin x}$.
Using half-angle identities $1+\cos x = 2\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$:
$-\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = -\cot\frac{x}{2}$.
Thus,$y = \cot^{-1}(-\cot\frac{x}{2})$.
Since $\cot^{-1}(-z) = \pi - \cot^{-1}(z)$,we have $y = \pi - \cot^{-1}(\cot\frac{x}{2})$.
Given $x \in (0, \frac{\pi}{4})$,then $\frac{x}{2} \in (0, \frac{\pi}{8})$,so $y = \pi - \frac{x}{2}$.

(C) Let $ x = \sin ^{-1} \frac{3}{4} $. Then $\sin x = \frac{3}{4}$.
We know that $\sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4} = \frac{\pi}{2}$.
Therefore,the expression becomes $\cos \left[\sin ^{-1} \frac{3}{4} + (\sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4})\right]$.
$= \cos \left[\sin ^{-1} \frac{3}{4} + \frac{\pi}{2}\right]$.
Using the identity $\cos \left(\frac{\pi}{2} + \theta\right) = -\sin \theta$,we get:
$= -\sin \left(\sin ^{-1} \frac{3}{4}\right)$.
$= -\frac{3}{4}$.

(B) Let $ \theta = \cos ^{-1} \left(\frac{2}{\sqrt{5}}\right) $. Then $ \cos \theta = \frac{2}{\sqrt{5}} $.
We need to find $ \tan \left(\frac{\theta}{2}\right) $.
Using the half-angle formula,$ \tan \left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} $.
Substituting the value of $ \cos \theta $:
$ \tan \left(\frac{\theta}{2}\right) = \sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}} = \sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}} $.
Rationalizing the denominator:
$ \sqrt{\frac{(\sqrt{5}-2)(\sqrt{5}-2)}{(\sqrt{5}+2)(\sqrt{5}-2)}} = \sqrt{\frac{(\sqrt{5}-2)^2}{5-4}} = \sqrt{(\sqrt{5}-2)^2} = \sqrt{5}-2 $.
Thus,the value is $ \sqrt{5}-2 $.

(B) Given equation: $\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left( \frac{1}{3} \right)$
Since $\frac{\pi}{4} = \tan ^{-1}(1)$,we can write:
$\tan ^{-1} x = \tan ^{-1}(1) - \tan ^{-1} \left( \frac{1}{3} \right)$
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left( \frac{A - B}{1 + AB} \right)$:
$\tan ^{-1} x = \tan ^{-1} \left( \frac{1 - \frac{1}{3}}{1 + 1 \times \frac{1}{3}} \right)$
$\tan ^{-1} x = \tan ^{-1} \left( \frac{\frac{2}{3}}{\frac{4}{3}} \right)$
$\tan ^{-1} x = \tan ^{-1} \left( \frac{2}{4} \right)$
$\tan ^{-1} x = \tan ^{-1} \left( \frac{1}{2} \right)$
Therefore,$x = \frac{1}{2}$.

(C) We use the trigonometric identities $\sec ^2 \theta = 1 + \tan ^2 \theta$ and $\operatorname{cosec}^2 \theta = 1 + \cot ^2 \theta$.
Substituting $\theta = \tan ^{-1} 2$ and $\theta = \cot ^{-1} 3$ respectively:
$= (1 + \tan ^2(\tan ^{-1} 2)) + (1 + \cot ^2(\cot ^{-1} 3))$
$= (1 + (\tan(\tan ^{-1} 2))^2) + (1 + (\cot(\cot ^{-1} 3))^2)$
$= (1 + 2^2) + (1 + 3^2)$
$= (1 + 4) + (1 + 9)$
$= 5 + 10 = 15$.

(C) Given the equation: $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$.
We know the standard identities for $a \in (0, 1)$:
$\sin ^{-1}\left(\frac{2 a}{1+a^2}\right) = 2 \tan ^{-1} a$
$\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right) = 2 \tan ^{-1} a$
Substituting these into the given equation:
$2 \tan ^{-1} a + 2 \tan ^{-1} a = \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$
$4 \tan ^{-1} a = 2 \tan ^{-1} x$
$2 \tan ^{-1} a = \tan ^{-1} x$
Using the formula $2 \tan ^{-1} a = \tan ^{-1}\left(\frac{2 a}{1-a^2}\right)$:
$\tan ^{-1}\left(\frac{2 a}{1-a^2}\right) = \tan ^{-1} x$
Therefore,$x = \frac{2 a}{1-a^2}$.

(B) Let the given expression be $E = \cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$.
We can rewrite the expression as $E = \cos \left(\cos ^{-1} \frac{1}{5} + \left(\cos ^{-1} \frac{1}{5} + \sin ^{-1} \frac{1}{5}\right)\right)$.
Using the identity $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} \frac{1}{5} + \sin ^{-1} \frac{1}{5} = \frac{\pi}{2}$.
Substituting this into the expression,we get $E = \cos \left(\cos ^{-1} \frac{1}{5} + \frac{\pi}{2}\right)$.
Using the trigonometric identity $\cos \left(\theta + \frac{\pi}{2}\right) = -\sin \theta$,we have $E = -\sin \left(\cos ^{-1} \frac{1}{5}\right)$.
Let $\theta = \cos ^{-1} \frac{1}{5}$,then $\cos \theta = \frac{1}{5}$.
Since $\sin \theta = \sqrt{1 - \cos^2 \theta}$,we have $\sin \theta = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2 \sqrt{6}}{5}$.
Therefore,$E = -\frac{2 \sqrt{6}}{5}$.