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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