Question

Prove that :
${sin}^{-1}\frac{3}{5}-{cos}^{-1}\frac{12}{13}={sin}^{-1}\frac{16}{65}$.

Answer 1

$L.H.S={\sin }^{-1}\left(\frac{3}{5}\right)-{\cos }^{-1}\left(\frac{12}{13}\right)$

$={\sin }^{-1}\left(\frac{3}{5}\right)-{\sin }^{-1}\left(\sqrt{-1{\left(\frac{12}{13}\right)}^2}\right)$

$[\because {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}]$

$={\sin }^{-1}\left(\frac{3}{5}\right)-{\sin }^{-1}\left(\frac{5}{13}\right)$

$={\sin }^{-1}[\frac{3}{5}\sqrt{1-\frac{25}{169}}-\frac{5}{13}\sqrt{1-\frac{9}{25}}]$

$[\because {\sin }^{-1}x-{\sin }^{-1}y={\sin }^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2}\left)\right]$

$={\sin }^{-1}[\frac{3}{5}\times \frac{12}{13}-\frac{5}{13}\times \frac{4}{5}]$

${\sin }^{-1}\left(\frac{16}{65}\right)=R.H.S.$ Hence proved.