If $\cos \,(\,\alpha + \,\beta )\, = \,\frac{3}{5}$ then $\tan \,(\,\alpha + \,\beta )\, = \,\frac{3}{4}$ and if $\sin \,(\,\alpha - \,\beta )\, = \,\frac{5}{{13}}$ then $\tan \,(\,\alpha - \,\beta )\, = \,\frac{5}{{12}}$
(since $\alpha - \,\beta $ here lies in the first quadrant)
Now $\tan \,(\,2\alpha ) = \tan \{ (\alpha \, + \,\beta ) + (\alpha \, - \,\beta )\} $
$ = \,\frac{{\tan \,(\alpha \, + \,\beta ) + \tan \,(\alpha \, - \,\beta )}}{{1 - \tan \,(\alpha \, + \,\beta ).\tan \,(\alpha \, - \,\beta )}}$
$ = \frac{{\frac{4}{3} + \frac{5}{{12}}}}{{1 - \frac{4}{3}.\frac{5}{{12}}}} = \frac{{63}}{{16}}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\sin ^{-1}\left(\sum_{i=1}^{\infty} x^{i+1}-x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^i\right)=\frac{\pi}{2}-\cos ^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^i-\sum_{i=1}^{\infty}(-x)^i\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is. . . . .
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[0, \pi]$, respectively.)