Question

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

Answer 1

Let $x={\sin }^{-1}\left(\frac{3}{5}\right)$
$\Rightarrow \sin x=\frac{3}{5}$
Now,
$\cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
Thus,
$\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}$
So, $x={\tan }^{-1}\frac{3}{4}$
i.e., ${\sin }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\frac{3}{4}$
Now, from given equation
$L.H.S.=2{\sin }^{-1}\frac{3}{5}$
$=2{\tan }^{-1}\left(\frac{3}{4}\right)$
Using $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$
$2{\tan }^{-1}\left(\frac{3}{4}\right)={\tan }^{-1}\left(\frac{2\left(\frac{3}{4}\right)}{1-{\left(\frac{3}{4}\right)}^2}\right)$
$={\tan }^{-1}\left(\frac{24}{7}\right)=R.H.S.$
Hence proved.