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Question:

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

(a) 0

(b) 1

(c) $-1$

(d) $\frac{1}{2}$

Solution:

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

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

$\Rightarrow 2 \tan ^{-1} x+\frac{\pi}{2}=\pi$                        $\left[\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]$

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

$\Rightarrow \tan ^{-1} x=\frac{\pi}{4}$

$\Rightarrow x=\tan \frac{\pi}{4}=1$

Thus, the value of $x$ is 1 .

Hence, the correct answer is option (b).

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