Question

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

Answer 1

Let ${\sin }^{-1}\frac{3}{5}=x\Rightarrow \sin x=\frac{3}{5}$.

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

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

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

Now, we have:
$L.H.S.=2{\sin }^{-1}\frac{3}{5}=2{\tan }^{-1}\frac{3}{4}$

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

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

$={\tan }^{-1}\frac{24}{7}$

$=R.H.S.$