If $\sin ^{-1} x=y,$ then

Considering only the principal values of an inverse function,the set $A = \{x \geq 0 \mid \tan^{-1} x + \tan^{-1} 6x = \frac{\pi}{4}\}$

The interval for which ${\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}$ holds is:

Consider the following statements:
Statement-$I$: $\operatorname{Cosh}^{-1} x = \operatorname{Tanh}^{-1} x$ has no solution.
Statement-$II$: $\operatorname{Cosh}^{-1} x = \operatorname{Coth}^{-1} x$ has only one solution.
The correct answer is:

If $y = \tan^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right]$,then $\frac{dy}{dx} = $

The value of $\sin^{-1}\left(\sin \frac{7 \pi}{6}\right)$ is equal to . . . . . . .