Question
Prove that: $2{\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{24}}{7}$

Answer

Let ${\sin ^{ - 1}}\frac{3}{5} = \theta$ so that $\sin \theta = \frac{3}{5}$
$\therefore \;\cos \theta = \sqrt {1 - {{\sin }^2}\theta } = \sqrt {1 - \frac{9}{{25}}} = \sqrt {\frac{{16}}{{25}}} = \frac{4}{5}$
$\therefore \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{3}{4}$
Since $\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$ = \frac{{2 \times \frac{3}{4}}}{{1 - \frac{9}{{16}}}} = \frac{{\frac{3}{2}}}{{\frac{7}{{16}}}} = \frac{{24}}{7}$
$\Rightarrow 2\theta = {\tan ^{ - 1}}\frac{{24}}{7}$
$ \Rightarrow 2{\sin ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{24}}{7}$

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