If $\alpha$ and $\beta$ are angles in the first quadrant such that $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$,then $\alpha + 2\beta =$ (in $^{\circ}$)

If $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,then $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ is equal to $........$.

If $\angle A = 90^\circ$ in the triangle $ABC$,then $\tan^{-1}\left(\frac{c}{a+b}\right) + \tan^{-1}\left(\frac{b}{a+c}\right) = $

If $\sin^{-1} x + \sin^{-1} y = \frac{2\pi}{3}$,then $\cos^{-1} x + \cos^{-1} y = $

Evaluate: ${\tan ^{ - 1}}1 + {\tan ^{ - 1}}2 + {\tan ^{ - 1}}3$

If for $x \in \left(0, \frac{1}{4}\right)$,the derivative of $\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$ is $\sqrt{x} \cdot g(x)$,then $g(x)$ equals