Question

Solve the equation $3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

Answer 1

• we have,

$3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

Put $x=\tan \theta$ then, $\theta ={\tan }^{-1}x$

$3{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-4{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+2{\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)=\frac{\pi }{3}$

$3{\sin }^{-1}\sin 2\theta -4{\cos }^{-1}\cos 2\theta +2{\tan }^{-1}\tan 2\theta =\frac{\pi }{3}$

$6\theta -8\theta +4\theta =\frac{\pi }{3}$

$2\theta =\frac{\pi }{3}$

$\theta =\frac{\pi }{3}$

Put $\theta ={\tan }^{-1}x$

${\tan }^{-1}x=\frac{\pi }{6}$

$x=\tan \frac{\pi }{6}$

$x=\frac{1}{\sqrt{3}}$

Hence, this is the answer.