If $3{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \pi$, then $x$ equals to

Given that, $3{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \pi$

…..(i)
$\Rightarrow$ $2{\tan ^{ - 1}}x + {\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \pi$
$\Rightarrow$ $2{\tan ^{ - 1}}x = \pi - \frac{\pi }{2}$

$\Rightarrow$ $2{\tan ^{ - 1}}x = \frac{\pi }{2}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{2}$

$\Rightarrow$ $\frac{{2x}}{{1 - {x^2}}} = \tan \frac{\pi }{2}$

$\Rightarrow$ $\frac{{2x}}{{1 - {x^2}}} = \frac{1}{0} \Rightarrow 1 - {x^2} = 0$

$\Rightarrow$ ${x^2} = 1 \Rightarrow x = \pm 1 \Rightarrow x = 1$

Hence, only $x = 1$ satisfies the given equation.

Note Here, putting $x = - 1$ in the given equation,

we get
$3{\tan ^{ - 1}}( - 1) + {\cot ^{ - 1}}( - 1) = \pi$

$\Rightarrow$ $3{\tan ^{ - 1}}\left[ {\tan \left( {\frac{{ - \pi }}{4}} \right)} \right] + {\cot ^{ - 1}}\left[ {\cot \left( {\frac{{ - \pi }}{4}} \right)} \right] = \pi$

$\Rightarrow$ $3{\tan ^{ - 1}}\left( { - \tan \frac{\pi }{4}} \right) + {\cot ^{ - 1}}\left( { - \cot \frac{\pi }{4}} \right) = \pi$

$\Rightarrow$ $- 3{\tan ^{ - 1}}\left( {\tan \frac{\pi }{4}} \right) + \pi - {\cot ^{ - 1}}\left( {\cot \frac{\pi }{4}} \right) = \pi$

$\Rightarrow$ $- 3 \cdot \frac{\pi }{4} + \pi - \frac{\pi }{4} = \pi$

$\Rightarrow$ $- \pi + \pi = \pi \Rightarrow 0 \ne \pi$

Hence, $x = - 1$ does not satisfy the given equation.