The value of $\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2})$ is equal to

If $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,then $\cos^{-1}(\sin \theta)$ is equal to

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:

The value of $\sin ^{-1}\left(-\frac{1}{2}\right)+\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is

If $e^{\left(\sinh ^{-1} 2+\cosh ^{-1} \sqrt{6}\right)}=(a+(b+\sqrt{c}) \sqrt{a}+b \sqrt{c})$,then $a+b+c=$

If $2\cos^{-1}\sqrt{\frac{1+x}{2}} = \frac{\pi}{2},$ then $x = $