Basic Concepts of ITF Questions in English

(B) Given the inequality: $\tan^{-1} \frac{x}{\pi} < \frac{\pi}{3}$.
Since the function $f(t) = \tan(t)$ is strictly increasing for $t \in (-\frac{\pi}{2}, \frac{\pi}{2})$,we can take the tangent of both sides:
$\tan(\tan^{-1} \frac{x}{\pi}) < \tan(\frac{\pi}{3})$
$\frac{x}{\pi} < \sqrt{3}$
$x < \sqrt{3} \times \pi$
Using the approximation $\sqrt{3} \approx 1.732$ and $\pi \approx 3.14159$:
$x < 1.732 \times 3.14159 \approx 5.441$
Since $x \in N$ (the set of natural numbers),the largest integer $x$ satisfying $x < 5.441$ is $5$.
Therefore,the maximum value of $x$ is $5$.

(C) Let $\cos^{-1} \left( \frac{\sqrt{5}}{3} \right) = \alpha$.
Then $\cos \alpha = \frac{\sqrt{5}}{3}$,where $0 < \alpha < \frac{\pi}{2}$.
We need to find $\tan \left( \frac{\alpha}{2} \right)$.
Using the half-angle formula,$\tan \left( \frac{\alpha}{2} \right) = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}$.
Substituting the value of $\cos \alpha$:
$\tan \left( \frac{\alpha}{2} \right) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}} = \sqrt{\frac{3 - \sqrt{5}}{3 + \sqrt{5}}}$.
To rationalize the denominator,multiply the numerator and denominator by $(3 - \sqrt{5})$:
$\sqrt{\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}} = \sqrt{\frac{(3 - \sqrt{5})^2}{9 - 5}} = \sqrt{\frac{(3 - \sqrt{5})^2}{4}} = \frac{3 - \sqrt{5}}{2}$.

(C) The principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{7\pi}{6}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we simplify the expression.
We know that $\sin(\pi + \theta) = -\sin \theta$,but it is easier to use $\sin(\pi + \theta) = \sin(2\pi - (\pi + \theta)) = \sin(\pi - \theta)$ is not correct here.
Instead,use $\sin(\frac{7\pi}{6}) = \sin(\pi + \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$.
Thus,$\sin^{-1}(\sin \frac{7\pi}{6}) = \sin^{-1}(\sin(-\frac{\pi}{6}))$.
Since $-\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the value is $-\frac{\pi}{6}$.

(C) Given equation: $\cos ^{-1} |x| + \cos ^{-1} |2x| = \pi$.
Since the range of $\cos ^{-1} y$ is $[0, \pi]$,for the sum to be $\pi$,we analyze the domain: $|x| \leq 1$ and $|2x| \leq 1$,which implies $|x| \leq 1/2$.
Case $1$: $x = 0$.
Substituting $x = 0$: $\cos ^{-1}(0) + \cos ^{-1}(0) = \pi/2 + \pi/2 = \pi$. Thus,$x = 0$ is a solution.
Case $2$: $x \neq 0$.
Let $\cos ^{-1} |x| = \alpha$ and $\cos ^{-1} |2x| = \beta$. Then $\alpha + \beta = \pi$,which implies $\beta = \pi - \alpha$.
Taking cosine on both sides: $\cos(\beta) = \cos(\pi - \alpha) = -\cos(\alpha)$.
Substituting back: $|2x| = -|x|$.
Since $|2x| \geq 0$ and $-|x| \leq 0$,this equality holds only if $|2x| = 0$ and $|x| = 0$,which leads back to $x = 0$.
Therefore,the only real solution is $x = 0$. The number of real solutions is $1$.

(C) We are given the equation $\sin^{-1} \theta = \sin^{-1}(\sin 5)$.
By the definition of the inverse sine function,$\sin^{-1}(\sin x) = x$ only if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Here,$x = 5$ radians. Since $\pi \approx 3.14$,$2\pi \approx 6.28$. Thus,$5$ is not in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] \approx [-1.57, 1.57]$.
We know that $\sin(x) = \sin(2\pi - x)$ is not correct,but $\sin(x) = \sin(x - 2\pi)$ is true because the period of the sine function is $2\pi$.
Specifically,$\sin(5) = \sin(5 - 2\pi)$.
Since $5 - 2\pi \approx 5 - 6.28 = -1.28$,which lies within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin^{-1}(\sin 5) = \sin^{-1}(\sin(5 - 2\pi)) = 5 - 2\pi$.
Therefore,$\theta = \sin(5 - 2\pi)$.

(C) We need to find the principal value of $\tan^{-1} \left( \cot \frac{43\pi}{4} \right)$.
First,simplify the argument of the cotangent function:
$\frac{43\pi}{4} = \frac{40\pi + 3\pi}{4} = 10\pi + \frac{3\pi}{4}$.
Since $\cot(n\pi + \theta) = \cot \theta$ for any integer $n$,we have:
$\cot \left( \frac{43\pi}{4} \right) = \cot \left( 10\pi + \frac{3\pi}{4} \right) = \cot \frac{3\pi}{4}$.
Now,express $\cot \theta$ in terms of $\tan$ using the identity $\cot \theta = \tan \left( \frac{\pi}{2} - \theta \right)$:
$\cot \frac{3\pi}{4} = \tan \left( \frac{\pi}{2} - \frac{3\pi}{4} \right) = \tan \left( \frac{2\pi - 3\pi}{4} \right) = \tan \left( -\frac{\pi}{4} \right)$.
Thus,$\tan^{-1} \left( \cot \frac{43\pi}{4} \right) = \tan^{-1} \left( \tan \left( -\frac{\pi}{4} \right) \right)$.
Since $-\frac{\pi}{4}$ lies in the principal value branch of $\tan^{-1}x$,which is $(-\frac{\pi}{2}, \frac{\pi}{2})$,the value is $-\frac{\pi}{4}$.

(A) Given the equation: $\sin(\cot^{-1}(1 + x)) = \cos(\tan^{-1}x)$.
Let $\theta = \cot^{-1}(1 + x)$,then $\cot \theta = 1 + x$. Since $\sin \theta = \frac{1}{\sqrt{1 + \cot^2 \theta}}$,we have $\sin(\cot^{-1}(1 + x)) = \frac{1}{\sqrt{1 + (1 + x)^2}}$.
Let $\phi = \tan^{-1}x$,then $\tan \phi = x$. Since $\cos \phi = \frac{1}{\sqrt{1 + \tan^2 \phi}}$,we have $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$.
Equating the two sides: $\frac{1}{\sqrt{1 + (1 + x)^2}} = \frac{1}{\sqrt{1 + x^2}}$.
Squaring both sides: $1 + (1 + x)^2 = 1 + x^2$.
$1 + 1 + 2x + x^2 = 1 + x^2$.
$2 + 2x = 1$.
$2x = -1$.
$x = -\frac{1}{2}$.

(D) Let $\theta = \cos^{-1} \sqrt{\frac{2}{3}}$.
Then $\cos \theta = \sqrt{\frac{2}{3}}$.
We know that $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
Substituting the value of $\cos \theta$,we get $\sin \theta = \sqrt{1 - \frac{2}{3}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.
Now,the expression becomes $\tan^{-1} (\sin \theta) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$.
Since $\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}$,we have $\tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}$.

(A) Let $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) = y$.
Then,$\sin y = \frac{1}{\sqrt{2}}$.
We know that the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Since $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,the principal value is $\frac{\pi}{4}$.

(B) Let $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right) = y$. Then,$\cot y = \frac{-1}{\sqrt{3}}$.
Since $\cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,we have $\cot y = -\cot \left(\frac{\pi}{3}\right)$.
Using the property $\cot(\pi - \theta) = -\cot \theta$,we get $\cot y = \cot \left(\pi - \frac{\pi}{3}\right) = \cot \left(\frac{2\pi}{3}\right)$.
The range of the principal value branch of $\cot^{-1}$ is $(0, \pi)$.
Since $\frac{2\pi}{3} \in (0, \pi)$,the principal value is $\frac{2\pi}{3}$.

(C) Let $y = \sin ^{-1}\left(-\frac{1}{2}\right)$.
Then,$\sin y = -\frac{1}{2}$.
We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,so $-\sin \left(\frac{\pi}{6}\right) = \sin \left(-\frac{\pi}{6}\right)$.
Thus,$\sin y = \sin \left(-\frac{\pi}{6}\right)$.
The range of the principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Since $-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,the principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi}{6}$.

(D) Let $y = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$.
Then,$\cos y = \frac{\sqrt{3}}{2}$.
We know that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.
The range of the principal value branch of $\cos^{-1} x$ is $[0, \pi]$.
Since $\frac{\pi}{6} \in [0, \pi]$,the principal value of $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is $\frac{\pi}{6}$.

(A) Let $cosec^{-1}(2) = y$.
Then,$cosec(y) = 2$.
We know that $cosec(\frac{\pi}{6}) = 2$.
Since the range of the principal value branch of $cosec^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$,and $\frac{\pi}{6}$ lies within this interval.
Therefore,the principal value of $cosec^{-1}(2)$ is $\frac{\pi}{6}$.

(B) Let $\tan ^{-1}(-\sqrt{3}) = y$.
Then,$\tan y = -\sqrt{3}$.
Since $\tan \frac{\pi}{3} = \sqrt{3}$,we have $\tan y = -\tan \frac{\pi}{3} = \tan \left(-\frac{\pi}{3}\right)$.
The range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Since $-\frac{\pi}{3} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,the principal value of $\tan ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$.

(C) Let $\cos ^{-1}\left(-\frac{1}{2}\right) = y$.
Then,$\cos y = -\frac{1}{2}$.
Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,we have $\cos y = -\cos \left(\frac{\pi}{3}\right)$.
Using the identity $\cos(\pi - \theta) = -\cos \theta$,we get $\cos y = \cos \left(\pi - \frac{\pi}{3}\right) = \cos \left(\frac{2 \pi}{3}\right)$.
The range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.
Since $\frac{2 \pi}{3} \in [0, \pi]$,the principal value of $\cos ^{-1}\left(-\frac{1}{2}\right)$ is $\frac{2 \pi}{3}$.

(D) Let $\tan ^{-1}(-1) = y$.
Then,$\tan y = -1$.
Since $\tan \left(\frac{\pi}{4}\right) = 1$,we have $\tan y = -\tan \left(\frac{\pi}{4}\right) = \tan \left(-\frac{\pi}{4}\right)$.
The range of the principal value branch of $\tan ^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Since $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,the principal value of $\tan ^{-1}(-1)$ is $-\frac{\pi}{4}$.

(A) Let $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right) = y$.
Then,$\sec y = \frac{2}{\sqrt{3}}$.
We know that $\sec \left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$.
The range of the principal value branch of $\sec ^{-1}$ is $[0, \pi] - \{\frac{\pi}{2}\}$.
Since $\frac{\pi}{6} \in [0, \pi] - \{\frac{\pi}{2}\}$,the principal value of $\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.

(B) Let $\cot ^{-1}(\sqrt{3}) = y$.
Then,$\cot y = \sqrt{3}$.
We know that $\cot \left(\frac{\pi}{6}\right) = \sqrt{3}$.
The range of the principal value branch of $\cot ^{-1}$ is $(0, \pi)$.
Since $\frac{\pi}{6} \in (0, \pi)$,the principal value of $\cot ^{-1}(\sqrt{3})$ is $\frac{\pi}{6}$.

(C) Let $y = \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.
Then $\cos y = -\frac{1}{\sqrt{2}}$.
Since $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,we have $\cos y = -\cos \left(\frac{\pi}{4}\right)$.
Using the identity $\cos (\pi - \theta) = -\cos \theta$,we get $\cos y = \cos \left(\pi - \frac{\pi}{4}\right) = \cos \left(\frac{3 \pi}{4}\right)$.
The range of the principal value branch of $\cos ^{-1}$ is $[0, \pi]$.
Since $\frac{3 \pi}{4} \in [0, \pi]$,the principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac{3 \pi}{4}$.

(D) Let $cosec^{-1}(-\sqrt{2}) = y$.
Then,$cosec\; y = -\sqrt{2}$.
We know that $cosec\left(\frac{\pi}{4}\right) = \sqrt{2}$,so $cosec\left(-\frac{\pi}{4}\right) = -\sqrt{2}$.
The range of the principal value branch of $cosec^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$.
Since $-\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}$,the principal value of $cosec^{-1}(-\sqrt{2})$ is $-\frac{\pi}{4}$.

(A) Let $\tan ^{-1}(1)=x$. Then,$\tan x=1=\tan \left(\frac{\pi}{4}\right)$. Thus,$\tan ^{-1}(1)=\frac{\pi}{4}$.
Let $\cos ^{-1}\left(-\frac{1}{2}\right)=y$. Then,$\cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)$. Thus,$\cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$.
Let $\sin ^{-1}\left(-\frac{1}{2}\right)=z$. Then,$\sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)$. Thus,$\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$.
Substituting these values,we get:
$\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}$.
Taking the least common multiple of the denominators $4, 3, 6$,which is $12$:
$= \frac{3 \pi + 8 \pi - 2 \pi}{12} = \frac{9 \pi}{12} = \frac{3 \pi}{4}$.

(B) Let $\cos ^{-1}\left(\frac{1}{2}\right)=x$.
Then,$\cos x=\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)$.
Therefore,$\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$.
Let $\sin ^{-1}\left(\frac{1}{2}\right)=y$.
Then,$\sin y=\frac{1}{2}=\sin \left(\frac{\pi}{6}\right)$.
Therefore,$\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$.
Substituting these values into the expression:
$\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}+2 \times \frac{\pi}{6} = \frac{\pi}{3}+\frac{\pi}{3} = \frac{2 \pi}{3}$.

(C) It is given that $\sin ^{-1} x=y$.
We know that the range of the principal value branch of the inverse sine function,$\sin ^{-1} x$,is defined as the closed interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Therefore,for any $x$ in the domain $[-1, 1]$,the value of $y$ must lie within this range.
Hence,$-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.

(D) Let $\tan ^{-1} \sqrt{3} = x$.
Then,$\tan x = \sqrt{3} = \tan \frac{\pi}{3}$.
We know that the range of the principal value branch of $\tan ^{-1}$ is $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.
Therefore,$\tan ^{-1} \sqrt{3} = \frac{\pi}{3}$.
Let $\sec ^{-1}(-2) = y$.
Then,$\sec y = -2 = -\sec \left( \frac{\pi}{3} \right) = \sec \left( \pi - \frac{\pi}{3} \right) = \sec \frac{2 \pi}{3}$.
We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi] - \left\{ \frac{\pi}{2} \right\}$.
Therefore,$\sec ^{-1}(-2) = \frac{2 \pi}{3}$.
Thus,$\tan ^{-1}(\sqrt{3}) - \sec ^{-1}(-2) = \frac{\pi}{3} - \frac{2 \pi}{3} = -\frac{\pi}{3}$.

(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 simplest form as $\sec ^{-1} x$.

(B) Given expression: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$
Using trigonometric identities $1 - \cos x = 2 \sin^2 \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$,we get:
$= \tan ^{-1}\left(\sqrt{\frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}}}\right)$
$= \tan ^{-1}\left(\sqrt{\tan^2 \frac{x}{2}}\right)$
$= \tan ^{-1}\left(\left| \tan \frac{x}{2} \right|\right)$
Since $x < \pi$,we have $\frac{x}{2} < \frac{\pi}{2}$. Assuming $x > 0$,$\tan \frac{x}{2}$ is positive,so $\left| \tan \frac{x}{2} \right| = \tan \frac{x}{2}$.
$= \tan ^{-1}\left(\tan \frac{x}{2}\right) = \frac{x}{2}$

(C) Given expression: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
Divide the numerator and denominator by $\cos x$:
$= \tan ^{-1}\left(\frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}\right)$
$= \tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
Using the formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$,where $A = \frac{\pi}{4}$ and $B = x$:
$= \tan ^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right)$
$= \frac{\pi}{4}-x$

(D) Given expression: $\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}$
Let $x = a \sin \theta$. Then $\frac{x}{a} = \sin \theta$,which implies $\theta = \sin ^{-1} \left(\frac{x}{a}\right)$.
Substituting $x = a \sin \theta$ into the expression:
$= \tan ^{-1} \left(\frac{a \sin \theta}{\sqrt{a^{2} - a^{2} \sin ^{2} \theta}}\right)$
$= \tan ^{-1} \left(\frac{a \sin \theta}{\sqrt{a^{2}(1 - \sin ^{2} \theta)}}\right)$
$= \tan ^{-1} \left(\frac{a \sin \theta}{a \cos \theta}\right)$
$= \tan ^{-1} (\tan \theta)$
$= \theta = \sin ^{-1} \left(\frac{x}{a}\right)$

(B) Let $\sin ^{-1} \left(\frac{1}{2}\right) = x$.
Then,$\sin x = \frac{1}{2} = \sin \left(\frac{\pi}{6}\right)$.
Therefore,$\sin ^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$.
Now,substitute this value into the expression:
$\tan ^{-1} \left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right] = \tan ^{-1} \left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$.
$= \tan ^{-1} \left[2 \cos \left(\frac{\pi}{3}\right)\right]$.
Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,we have:
$= \tan ^{-1} \left[2 \times \frac{1}{2}\right] = \tan ^{-1} (1)$.
Since $\tan \left(\frac{\pi}{4}\right) = 1$,the value is $\frac{\pi}{4}$.

(C) We need to evaluate $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$.
The principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Since $\frac{2 \pi}{3} \approx 120^\circ$,which is not in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we use the property $\sin(\pi - x) = \sin x$.
Therefore,$\sin \frac{2 \pi}{3} = \sin \left(\pi - \frac{2 \pi}{3}\right) = \sin \frac{\pi}{3}$.
Now,$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) = \sin ^{-1}\left(\sin \frac{\pi}{3}\right)$.
Since $\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we have $\sin ^{-1}\left(\sin \frac{\pi}{3}\right) = \frac{\pi}{3}$.
Thus,the correct value is $\frac{\pi}{3}$.

(D) We need to evaluate $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$.
The principal value branch of $\tan ^{-1} x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Since $\frac{3 \pi}{4} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,we use the property $\tan(\pi - \theta) = -\tan \theta$.
Thus,$\tan \frac{3 \pi}{4} = \tan \left(\pi - \frac{\pi}{4}\right) = -\tan \frac{\pi}{4} = \tan \left(-\frac{\pi}{4}\right)$.
Now,$\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right) = \tan ^{-1}\left(\tan \left(-\frac{\pi}{4}\right)\right)$.
Since $-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,we have $\tan ^{-1}\left(\tan \left(-\frac{\pi}{4}\right)\right) = -\frac{\pi}{4}$.

(B) We know that $\cos ^{-1}(\cos x) = x$ if $x \in [0, \pi]$,which is the principal value branch of $\cos ^{-1} x$.
Here,$\frac{7 \pi}{6} \notin [0, \pi]$.
We use the property $\cos(2\pi - \theta) = \cos \theta$ to bring the angle within the principal range.
$\cos \left(\frac{7 \pi}{6}\right) = \cos \left(2 \pi - \frac{7 \pi}{6}\right) = \cos \left(\frac{5 \pi}{6}\right)$.
Since $\frac{5 \pi}{6} \in [0, \pi]$,we have:
$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right) = \cos ^{-1}\left(\cos \frac{5 \pi}{6}\right) = \frac{5 \pi}{6}$.

(C) Let $\sin ^{-1}\left(-\frac{1}{2}\right) = x$.
Then,$\sin x = -\frac{1}{2}$.
Since the range of the principal value branch of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we have $\sin ^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$.
Now,substitute this value into the expression:
$\sin \left(\frac{\pi}{3} - \left(-\frac{\pi}{6}\right)\right) = \sin \left(\frac{\pi}{3} + \frac{\pi}{6}\right)$.
$= \sin \left(\frac{2\pi + \pi}{6}\right) = \sin \left(\frac{3\pi}{6}\right) = \sin \left(\frac{\pi}{2}\right)$.
Since $\sin \left(\frac{\pi}{2}\right) = 1$,the final value is $1$.

(D) We know that the principal value branch of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{3 \pi}{5} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$,we cannot directly write $\sin ^{-1}(\sin x) = x$.
We use the property $\sin(\pi - \theta) = \sin \theta$.
Thus,$\sin(\frac{3 \pi}{5}) = \sin(\pi - \frac{3 \pi}{5}) = \sin(\frac{2 \pi}{5})$.
Since $\frac{2 \pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin ^{-1}(\sin \frac{2 \pi}{5}) = \frac{2 \pi}{5}$.
Therefore,$\sin ^{-1}(\sin \frac{3 \pi}{5}) = \frac{2 \pi}{5}$.

(D) Given the equation $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan ^{-1} \left( \frac{2x+3x}{1-(2x)(3x)} \right) = \frac{\pi}{4}$
$\tan ^{-1} \left( \frac{5x}{1-6x^2} \right) = \frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{5x}{1-6x^2} = \tan \left( \frac{\pi}{4} \right) = 1$
$5x = 1 - 6x^2$
$6x^2 + 5x - 1 = 0$
$(6x - 1)(x + 1) = 0$
Thus,$x = \frac{1}{6}$ or $x = -1$.
If $x = -1$,then $\tan ^{-1}(-2) + \tan ^{-1}(-3) = -(\tan ^{-1} 2 + \tan ^{-1} 3)$,which is negative and not equal to $\frac{\pi}{4}$.
Therefore,the only valid solution is $x = \frac{1}{6}$.

(A) We know that $\cos ^{-1}(\cos x) = x$ if $x \in [0, \pi]$,which is the principal value branch of $\cos ^{-1} x$.
Here,$\frac{13 \pi}{6} \notin [0, \pi]$.
Now,$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ can be written as:
$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) = \cos ^{-1}\left[\cos \left(2 \pi + \frac{\pi}{6}\right)\right]$.
Since $\cos(2 \pi + \theta) = \cos \theta$,we have:
$\cos ^{-1}\left[\cos \left(2 \pi + \frac{\pi}{6}\right)\right] = \cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]$.
Since $\frac{\pi}{6} \in [0, \pi]$,we get:
$\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right] = \frac{\pi}{6}$.

(B) We know that $\tan ^{-1}(\tan x) = x$ only if $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,which is the principal value branch of $\tan ^{-1} x$.
Here,the given value is $\frac{7 \pi}{6}$,and $\frac{7 \pi}{6} \notin \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
We can simplify the expression using the periodicity of the tangent function: $\tan \left(\pi + \theta\right) = \tan \theta$.
Therefore,$\tan \left(\frac{7 \pi}{6}\right) = \tan \left(\pi + \frac{\pi}{6}\right) = \tan \left(\frac{\pi}{6}\right)$.
Since $\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,we have $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) = \tan ^{-1}\left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6}$.

(C) Let $\tan^{-1} x = y$. Then $\tan y = x$.
Since $\tan y = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1}$,we can consider a right-angled triangle where the opposite side is $x$ and the adjacent side is $1$.
The hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.
Therefore,$\sin y = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$.
Thus,$\sin(\tan^{-1} x) = \sin y = \frac{x}{\sqrt{1+x^2}}$.

(D) We use the Taylor series expansions for $\sin^{-1} x$ and $\tan^{-1} x$ near $x = 0$:
$\sin^{-1} x = x + \frac{x^3}{6} + O(x^5)$
$\tan^{-1} x = x - \frac{x^3}{3} + O(x^5)$
Substituting these into the limit expression:
$L = \lim_{x \rightarrow 0} \frac{(x + \frac{x^3}{6}) - (x - \frac{x^3}{3})}{3x^3}$
$L = \lim_{x \rightarrow 0} \frac{\frac{x^3}{6} + \frac{x^3}{3}}{3x^3} = \lim_{x \rightarrow 0} \frac{\frac{1}{6} + \frac{1}{3}}{3} = \frac{\frac{1+2}{6}}{3} = \frac{3/6}{3} = \frac{1/2}{3} = \frac{1}{6}$
Given $L = 1/6$,we calculate $6L + 1$:
$6(1/6) + 1 = 1 + 1 = 2$

(B) Given expression: $\tan ^{-1}\left(\frac{\cos \left(\frac{15 \pi}{4}\right)-1}{\sin \left(\frac{\pi}{4}\right)}\right)$
Since $\frac{15 \pi}{4} = 4 \pi - \frac{\pi}{4}$,we have $\cos \left(\frac{15 \pi}{4}\right) = \cos \left(4 \pi - \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$.
Substituting this into the expression: $\tan ^{-1}\left(\frac{\frac{1}{\sqrt{2}}-1}{\frac{1}{\sqrt{2}}}\right) = \tan ^{-1}\left(\frac{1-\sqrt{2}}{1}\right) = \tan ^{-1}(1-\sqrt{2})$.
Using the identity $\tan \left(\frac{\theta}{2}\right) = \frac{1-\cos \theta}{\sin \theta}$,we have $\tan \left(\frac{\pi}{8}\right) = \sqrt{2}-1$.
Therefore,$\tan \left(-\frac{\pi}{8}\right) = -(\sqrt{2}-1) = 1-\sqrt{2}$.
Thus,$\tan ^{-1}(1-\sqrt{2}) = -\frac{\pi}{8}$.

(C) Let $\tan ^{-1} \left(\frac{4}{3}\right) = \theta$,which implies $\tan \theta = \frac{4}{3}$.
From the right-angled triangle with opposite side $4$ and adjacent side $3$,the hypotenuse is $\sqrt{3^2 + 4^2} = 5$.
Thus,$\cos \theta = \frac{3}{5}$ and $\sin \theta = \frac{4}{5}$.
Let the given expression be $E = \cos ^{-1}\left(\frac{3}{10} \cos \theta + \frac{2}{5} \sin \theta\right)$.
Substituting the values of $\cos \theta$ and $\sin \theta$:
$E = \cos ^{-1}\left(\frac{3}{10} \times \frac{3}{5} + \frac{2}{5} \times \frac{4}{5}\right)$
$E = \cos ^{-1}\left(\frac{9}{50} + \frac{8}{25}\right)$
$E = \cos ^{-1}\left(\frac{9}{50} + \frac{16}{50}\right)$
$E = \cos ^{-1}\left(\frac{25}{50}\right)$
$E = \cos ^{-1}\left(\frac{1}{2}\right)$
Since the principal value of $\cos ^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi}{3}$,the final answer is $\frac{\pi}{3}$.

(C) Let $x = \tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)$.
Simplify the argument: $\frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$.
So,$x = \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$.
Let $y = \sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)$.
Simplify the argument: $\sqrt{\frac{4(2+\sqrt{3})}{3(2+\sqrt{3})}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.
So,$y = \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$.
Therefore,$x + y = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

(C) Given the function $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$.
To check for odd/even,we evaluate $g(-u)$:
$g(-u) = 2 \tan^{-1}(e^{-u}) - \frac{\pi}{2}$.
Using the identity $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we know $\tan^{-1}(e^{-u}) = \frac{\pi}{2} - \tan^{-1}(e^u)$.
Substituting this into $g(-u)$:
$g(-u) = 2 \left( \frac{\pi}{2} - \tan^{-1}(e^u) \right) - \frac{\pi}{2} = \pi - 2 \tan^{-1}(e^u) - \frac{\pi}{2} = \frac{\pi}{2} - 2 \tan^{-1}(e^u) = -g(u)$.
Since $g(-u) = -g(u)$,the function is odd.
To check for monotonicity,we find the derivative $g'(u)$:
$g'(u) = \frac{d}{du} (2 \tan^{-1}(e^u) - \frac{\pi}{2}) = 2 \cdot \frac{1}{1 + (e^u)^2} \cdot e^u = \frac{2e^u}{1 + e^{2u}}$.
Since $e^u > 0$ for all $u \in (-\infty, \infty)$,$g'(u) > 0$ for all $u$.
Therefore,$g$ is strictly increasing in $(-\infty, \infty)$.

(A) Let $S = \sum_{k=0}^{10} \sec \left(\frac{7 \pi}{12}+\frac{k \pi}{2}\right) \sec \left(\frac{7 \pi}{12}+\frac{(k+1) \pi}{2}\right)$.
Using the identity $\sec A \sec B = \frac{\sin(B-A)}{\cos A \cos B \sin(B-A)}$,we note that $B-A = \frac{\pi}{2}$.
Thus,$\sec A \sec B = \frac{\sin(\pi/2)}{\cos A \cos B \sin(\pi/2)} = \frac{\tan B - \tan A}{\sin(\pi/2)} = \tan B - \tan A$.
Here,$A = \frac{7 \pi}{12} + \frac{k \pi}{2}$ and $B = \frac{7 \pi}{12} + \frac{(k+1) \pi}{2}$.
So,the sum becomes $\sum_{k=0}^{10} (\tan(\frac{7 \pi}{12} + \frac{(k+1) \pi}{2}) - \tan(\frac{7 \pi}{12} + \frac{k \pi}{2}))$.
This is a telescoping sum: $\tan(\frac{7 \pi}{12} + \frac{11 \pi}{2}) - \tan(\frac{7 \pi}{12})$.
Since $\tan(\theta + \frac{11 \pi}{2}) = \tan(\theta - \frac{\pi}{2}) = -\cot \theta$,the sum is $-\cot(\frac{7 \pi}{12}) - \tan(\frac{7 \pi}{12}) = -(\frac{\cos(7 \pi / 12)}{\sin(7 \pi / 12)} + \frac{\sin(7 \pi / 12)}{\cos(7 \pi / 12)}) = -\frac{1}{\sin(7 \pi / 12) \cos(7 \pi / 12)} = -\frac{2}{\sin(7 \pi / 6)} = -\frac{2}{-1/2} = 4$.
The expression is $\sec^{-1}(\frac{1}{4} \times 4) = \sec^{-1}(1) = 0$.

(A) We are given the expression $\cot ^{-1}\left(2 \cos \left(2 \operatorname{cosec}^{-1}(\sqrt{2})\right)\right)$.
First,evaluate $\operatorname{cosec}^{-1}(\sqrt{2})$.
Since $\operatorname{cosec}(\frac{\pi}{4}) = \sqrt{2}$,we have $\operatorname{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$.
Now,substitute this into the expression:
$2 \cos \left(2 \times \frac{\pi}{4}\right) = 2 \cos \left(\frac{\pi}{2}\right)$.
Since $\cos \left(\frac{\pi}{2}\right) = 0$,the expression becomes $2 \times 0 = 0$.
Finally,we need to find $\cot ^{-1}(0)$.
Since $\cot \left(\frac{\pi}{2}\right) = 0$,we have $\cot ^{-1}(0) = \frac{\pi}{2}$.
Thus,the correct option is $A$.

(A) Let $\cot ^{-1} x = t$.
Then,$x = \cot t$.
We know the identity $1 + \cot^2 t = \operatorname{cosec}^2 t$.
Substituting $x$,we get $1 + x^2 = \operatorname{cosec}^2 t$.
Therefore,$\operatorname{cosec} t = \sqrt{1 + x^2}$.
Since $\sin t = \frac{1}{\operatorname{cosec} t}$,we have $\sin t = \frac{1}{\sqrt{1 + x^2}}$.
Thus,$\sin (\cot ^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$.

(A) We know that the range of the principal value branch of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{3 \pi}{4}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we simplify the expression:
$\sin ^{-1}\left(\sin \frac{3 \pi}{4}\right) = \sin ^{-1}\left[\sin \left(\pi - \frac{\pi}{4}\right)\right]$
Using the identity $\sin(\pi - \theta) = \sin \theta$,we get:
$= \sin ^{-1}\left(\sin \frac{\pi}{4}\right)$
Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the expression simplifies to:
$= \frac{\pi}{4}$

(C) Let $\theta = \sin ^{-1} 0.8$. Then $\sin \theta = 0.8$.
We need to find the value of $\sin(2\theta)$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we first find $\cos \theta$.
Since $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (0.8)^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$.
Now,substitute the values into the identity:
$\sin(2\theta) = 2 \times 0.8 \times 0.6 = 0.96$.

(C) We know that the range of the principal value branch of $\sin ^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{2 \pi}{3}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we use the property $\sin(\pi - \theta) = \sin(\theta)$.
$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right) = \sin ^{-1}\left(\sin \left(\pi - \frac{\pi}{3}\right)\right)$
$= \sin ^{-1}\left(\sin \left(\frac{\pi}{3}\right)\right)$
$= \frac{\pi}{3}$