Question

lf ${\tan }^{-1}(x+\frac{2}{x})$ $-$ ${\tan }^{-1}(x-\frac{2}{x})$ $={\tan }^{-1}\left(\frac{4}{x}\right)$ then $x=$

• A. $\pm \sqrt{2}$
• B. $\pm 2$
• C. $\pm 1$
• D. $\pm \sqrt{3}$

Answer 1

The correct option is A $\pm \sqrt{2}$

$ta{n}^{-1}x-ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x-y}{1+xy}\right)x,y,>0$
$\therefore ta{n}^{-1}(x+\frac{2}{x})-ta{n}^{-1}(x-\frac{2}{x})=ta{n}^{-1}\left(\frac{\frac{4}{x}}{1+(x^2-\frac{4}{x^2})}\right)=ta{n}^{-1}\left(\frac{4}{x}\right)$ [Given]

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