Question

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

• A. $-1$
• B. $\frac{1}{6}$
• C. $-1,\frac{1}{6}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{6}$

Explanation for the correct option:

Step : Solve for roots.

$\begin{array}{c}{\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}\\ {\tan }^{-1}\frac{\left(2x+3x\right)}{\left(1-\left(2x\right)\left(3x\right)\right)}=\frac{\pi }{4}\\ {\tan }^{-1}\frac{\left(5x\right)}{\left(1-6x^2\right)}={\tan }^{-1}1\\ \frac{\left(5x\right)}{\left(1-6x^2\right)}=1\\ \left(1-6x^2\right)=5x\\ 6x^2+5x-1=0\\ \left(x+1\right)\left(x-\left(\frac{1}{6}\right)\right)=0\\ x=-1,\frac{1}{6}\end{array}$

But $-1$does not apply

Hence, option (B) is correct.