Question

If ${\tan }^{-1}(x+2)+{\tan }^{-1}(x-2)={\tan }^{-1}\left(\frac{1}{2}\right)$, than sum of value(s) of $x$ is equal to

• A. $1$
• B. $-5$
• C. $-4$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $-4$

Given:

$ta{n}^{-1}(x+2)+ta{n}^{-1}(x-2)=ta{n}^{-1}\left(\frac{1}{2}\right)$

$ta{n}^{-1}(x+2)+ta{n}^{-1}(x-2)=ta{n}^{-1}\left(\frac{x+2+x-2}{1-(x+2)(x-2)}\right)$

$=ta{n}^{-1}\left(\frac{2x}{1-x^2+4}\right)$

so,

$ta{n}^{-1}\left(\frac{2x}{1-x^2+4}\right)=ta{n}^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow \frac{2x}{1-x^2+4}=\frac{1}{2}$

$\Rightarrow 4x=1-x^2+4$

$\Rightarrow x^2+4x-5=0$

$\Rightarrow (x+5)(x-1)=0$

$\Rightarrow x=-5,1$

sum of -5 and 1 = -5 + 1 = -4