Question

If ${\tan }^{-1}\left(x+2\right)+{\tan }^{-1}\left(x-2\right)-{\tan }^{-1}\left(\frac{1}{2}\right)=0$ then one of the value of $x=$

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

Answer 1

The correct option is D

$1$

Explanation for the correct option:

Step1: Solve for $x$

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

Hence option (4) is correct