Let $0<\alpha<1, \beta=\frac{1}{3 \alpha}$ and $\tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)=\frac{\pi}{4}$. Then $6(\alpha+\beta)$ is equal to:

Considering the principal values of inverse trigonometric functions, the value of the expression

$$\tan \left( 2 \sin^{-1}\left( \frac{2}{\sqrt{13}} \right) - 2 \cos^{-1}\left( \frac{3}{\sqrt{10}} \right) \right)$$

is equal to :

If the domain of the function $f(x)=\sin ^{-1}\left(\frac{1}{x^2-2 x-2}\right)$, is $(-\infty, \alpha] \cup[\beta, \gamma] \cup[\delta, \infty)$, then $\alpha+\beta+\gamma+\delta$ is equal to

The number of solutions of $\tan ^{-1} 4 x+\tan ^{-1} 6 x=\frac{\pi}{6}$, where $-\frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}}$, is equal to :