Question

Find x:- ${\tan }^{-1}\left(2x\right)+{\tan }^{-1}\left(3x\right)=\pi /4$

Answer 1

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

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

$\frac{5x}{1-6x^2}=\tan \pi /4$

$5x=1-6x^2$

$6x^2+5x-1=0$

$(6x-1)(x+1)=0$

$x=\frac{1}{6},-1$