Question

Solve:-
${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$

Answer 1

${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}\Rightarrow {\tan }^{-1}\frac{2x+3x}{1-\left(6x^2\right)}=\frac{\pi }{4}\{{\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\frac{A+B}{1-AB}\}\Rightarrow 5x=(1-{6x}^2)\left(\tan \frac{\pi }{4}\right)\Rightarrow 5x=1-{6x}^2\Rightarrow {6x}^2+5x-1=0\Rightarrow {6x}^2+6x-x-1=0\Rightarrow 6x(x+1)-1(x+1)=0\Rightarrow (6x-1)(x+1)=0x=\frac{1}{6},-1$