Question

Prove that : $2{\sin }^{-1}\left(\frac{3}{5}\right)-{\tan }^{-1}\left(\frac{17}{31}\right)=\frac{\pi }{4}$

Answer 1

$Letusassume{sin}^{-1}\frac{3}{5}=t$

$\sin t=\frac{3}{5}{\sin }^{2}t=\frac{9}{25}1-{\cos }^{2}t=\frac{9}{25}{\cos }^{2}t=\frac{16}{25}\cos t=\frac{4}{5}\tan t=\frac{\sin t}{\cos t}\tan t=\frac{3}{5}t={\tan }^{-1}\frac{3}{5}now2{\sin }^{-1}\frac{3}{5}=2{\tan }^{-1}\frac{3}{5}2{tan}^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}thereforbyputtinginaboveformula2{\tan }^{-1}\frac{3}{5}={\tan }^{-1}\frac{24}{7}$

$Now,accordingtoquestion{\tan }^{-1}\frac{24}{7}-{\tan }^{-1}\frac{17}{31}---\left(1\right){\tan }^{-1}x={\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)nowapplytheaboveformulainequation\left(1\right){\tan }^{-1}1nowwrite\tan \frac{\pi }{4}intheplaceofeq---\left(1\right){\tan }^{-1}\left(\tan \frac{\pi }{4}\right)=\frac{\pi }{4}$

Hence, proved.