Question

$If\alpha$ is two solution of equation ${\tan }^{-1}x+{\cot }^{-1}x=2{\tan }^{-1}x,then{\alpha }^2$ is equal to

Answer 1

${\tan }^{-1}x+{\cot }^{-1}x=2{\tan }^{-1}x$

${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=2{\tan }^{-1}x{\tan }^{-1}\left(\frac{x+\frac{1}{x}}{1-x\frac{1}{x}}\right)=2{\tan }^{-1}x$

$2{\tan }^{-1}x=\frac{\pi }{2}$

${\tan }^{-1}x=\frac{\pi }{4}$

$x=1$

$\alpha =1$

${\alpha }^2=1$