Question

If ${\tan }^{-1}x+{\tan }^{-1}2x+{\tan }^{-1}3x=\pi$, then :

• A. $x=0$
• B. $x=-1$
• C. $x=1$
• D. $xϵ\varphi$

Answer 1

Answer: (C)

$x=1$

Given, ${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(2x\right)=\pi -{\tan }^{-1}\left(3x\right)$
${\tan }^{-1}\left(\frac{3x}{1-2x^2}\right)=\pi -{\tan }^{-1}\left(3x\right)$
$\tan \left({\tan }^{-1}\right(\frac{3x}{1-2x^2}\left)\right)=\tan (\pi -{\tan }^{-1}(3x\left)\right)$
$\frac{3x}{1-2x^2}=-3x$
$2x^2-1=1$
$2x^2=2$
$x^2=1$
$x=\pm 1$
However, for $x=-1$, we get $-\pi$
Hence $x=1$ is the solution for the above equation.