Question

If ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$, then $x=$

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

Answer 1

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

Given that ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}⟹{\tan }^{-1}2x=\frac{\pi }{4}-{\tan }^{-1}3x⟹2x=\frac{1-3x}{1+3x}⟹2x+6x^2=1-3x⟹6x^2+5x-1=0⟹(6x-1)(x+1)=0⟹x=\frac{1}{6}orx=-1$

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

so,$x=\frac{1}{6}$ is the answer