Question

Number of real values of ' $x$ ' satisfying the equation $\tan ^{-1}\left(\frac{x-1}{x+1}\right)+\frac{\pi}{2}=\tan ^{-1}\left(4-2 x-2 x^2\right)+\cot ^{-1}\left(\frac{x-1}{x+3}\right)$ is

• A. 0
• B. 1
• C. 2
• D. more than 2

Answer 1

Answer: (C)

$\tan ^{-1}\left(\frac{x-1}{x+1}\right)+\tan ^{-1}\left(\frac{x-1}{x+3}\right)=\tan ^{-1}\left(4-2 x-2 x^2\right)$
$\Rightarrow \frac{\left(\frac{x-1}{x+1}\right)+\left(\frac{x-1}{x+3}\right)}{1-\frac{(x-1)^2}{(x+1)(x+3)}}=-2(x+2)(x-1)$
$\therefore x =1$ is a solution.
If $x \neq 1$, then $\frac{2(x+2)}{2(3 x+1)}=-2(x+2)$
$\therefore x =-2 \text { or } x =\frac{-1}{2} \text { (not is domain). }$