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

Answer

$\text{Let} \sin^{-1}\frac{3}{5}=x. \text{Then}, \sin x=\frac{3}{5}.$

$\Rightarrow\cos x=\sqrt{1-\bigg(\frac{3}{5}\bigg)^2}=\frac{4}{5}$

$\therefore\tan x=\frac{3}{4}$

$\therefore x=\tan^{-1}\frac{3}{4}\Rightarrow\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4}$

Now, we have:

$\text{L.H.S.}=2\sin^{-1}\frac{3}{5}=2\tan^{-1}\frac{3}{4}$

$=\tan^{-1}\Bigg(\frac{2\times\frac{3}{4}}{1-\left(\frac{3}{4}\right)^2}\Bigg)$ $\bigg[2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}\bigg]$

$=\tan^{-1}\Bigg(\frac{\frac{3}{2}}{\frac{16-9}{16}}\Bigg)=\tan^{-1}\bigg(\frac{3}{2}\times\frac{16}{7}\bigg)$

$=\tan^{-1}\frac{24}{7}=\text{R.H.S.}$

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