The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} 2a$ has a real solution, if

Considering only the principal values,if $\tan (\cos ^{ - 1}x) = \sin [\cot ^{ - 1}(1/2)]$,then $x$ is equal to

If $x_1, x_2, x_3$ are the roots of the equation $x^3 - x^2 \tan \theta + x \tan^2 \theta + \tan \theta = 0$ and $0 < \theta < \frac{\pi}{4}$, then the value of $\tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3$ at $\theta = \frac{\pi}{12}$ is

$\begin{aligned} & 2 \sin ^{-1} x+\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+3 \cos ^{-1} x \\ & -\cos ^{-1}\left(4 x^3-3 x\right) \text { is equal to }\end{aligned}$

$\cos \left(\cos ^{-1} \frac{1}{3}+\cos ^{-1} \frac{1}{5}\right)+\cos \left(\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{1}{5}\right) =$ . . . . . . .

If $\alpha$ and $\beta$ are the least and the greatest values of $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$ for all $x \in [-1, 1]$ respectively,then $8(\alpha+\beta)=$