Question:
If $\tan ^{-1}(\cot \theta)=2 \theta$, then $\theta=$
(a) $\pm \frac{\pi}{3}$
(b) $\pm \frac{\pi}{4}$
(c) $\pm \frac{\pi}{6}$
(d) none of these
Solution:
(c) $\pm \frac{\pi}{6}$
We have,
$\tan ^{-1}(\cot \theta)=2 \theta$
$\Rightarrow \tan 2 \theta=\cot \theta$
$\Rightarrow \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{1}{\tan \theta}$
$\Rightarrow 2 \tan ^{2} \theta=1-\tan ^{2} \theta$
$\Rightarrow 3 \tan ^{2} \theta=1$
$\Rightarrow \tan ^{2} \theta=\frac{1}{3}$
$\Rightarrow \tan \theta=\pm \frac{1}{\sqrt{3}}$
$\therefore \theta=\pm \frac{\pi}{6}$