Question

If ${\tan }^{-1}(x+\frac{3}{x})-{\tan }^{-1}(x-\frac{3}{x})={\tan }^{-1}\frac{6}{x}$, then the value of $x^4$is

Answer 1

$ta{n}^{-1}\left(\frac{x+\frac{3}{x}-x+\frac{3}{x}}{1+(x+\frac{3}{x})(x-\frac{3}{x})}\right)$

$=ta{n}^{-1}\left(\frac{\frac{6}{x}}{1+x^2-\frac{9}{x^2}}\right)$

$=ta{n}^{-1}\left(\frac{6}{x}\right)$

Then, $1+x^2-\frac{9}{x^2}=1$

$x^2-\frac{9}{x^2}=0$

$x^2=\frac{9}{x^2}$

$x^4=9$

$x=\pm \sqrt{3}$