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

$\begin{array}{1 1} 0 \\ 1 \\ -1 \\ \frac{1}{2}\end{array}$

Toolbox:
• $tan^{-1}x+cot^{-1}x=\large\frac{\pi}{2}$
• $tan^{-1}1=\large\frac{\pi}{4}$
Ans (B) 1

We can write the given equation as
$2tan^{-1}x+tan^{-1}x+cot^{-1}x=\large\frac{\pi}{2}$

$\Rightarrow\:2tan^{-1}x+\large\frac{\pi}{2}=\pi$
$\Rightarrow 2tan^{-1}x=\pi-\large\frac{\pi}{2}=\large\frac{\pi}{2}$
$\Rightarrow tan^{-1}x=\large\frac{\pi}{4}$

$\Rightarrow x = 1$

edited Mar 16, 2013