Question

The number of the solution $ta{n}^{-1}(1+x)+ta{n}^{-1}(1-x)=\frac{\pi }{2}$ is

• A. $2$
• B. $3$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

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

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

[$\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left[\frac{x+y}{1-xy}\right]$]

${\cot }^{-1}\left(x^2\right)=\frac{\pi }{2}$

Taking $\cot$ on both the sides, we get

$x^2=0$

$x=0$

Hence there is only one real solution for the above equation.