Question

A solution of the equation ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$ is

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D $\frac{1}{6}$

${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}\Rightarrow {\tan }^{-1}\frac{2x+3x}{1-2x\times 3x}=\frac{\pi }{4}\Rightarrow 5x=1\times (1-6x^2)\Rightarrow 6x^2+5x-1=0\Rightarrow (x+1)(6x-1)=0$

Hence the solution is $x=\frac{1}{6}$