Question

If ${\tan }^{-1}(x+2)+{\tan }^{-1}(x-2)-{\tan }^{-1}\left(\frac{1}{2}\right)=0,$ then one of the values of $x$ is equal to

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

Answer 1

Answer: (D)

Given that, ${\tan }^{-1}(x+2)+{\tan }^{-1}(x+2)-{\tan }^{-1}\left(\frac{1}{2}\right)=0$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{x+2+x-2}{1-(x^2-4)}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)=0$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{2x}{1-(x^2-4)}\right)-{\tan }^{-1}\left(\frac{1}{2}\right)=0$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{\left(\frac{2x}{1-(x^2-4)}-\frac{1}{2}\right)}{1+\frac{2x}{(1-(x^2-4))}.\frac{1}{2}}\right)=0$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{\frac{4x-1+x^2-4}{2(1-(x^2-4))}}{\frac{2(1-(x^2-4))+2x}{2(1-(x^2-4))}}\right)=0$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{x^2+4x-5}{-2x^2+2x+6}\right)=0$
$\Rightarrow$ $x^2+4x-5=0$
$\Rightarrow$ $x=1,-5$
$\Rightarrow$ $x=1$ $\because$ $-5$ is not possible.