Q86 / 200 · Manipal (MET) 2010 · Mathematics
$2 \tan ^{-1}\left(\operatorname{cosec} \tan ^{-1} x-\tan \cot ^{-1} x\right)$ is equal to
- A$\cot ^{-1} x$
- B$\cot ^{-1} \frac{1}{x}$
- Correct answerC$\tan ^{-1} x$
- DNone of these
Solution
The given expression
$\begin{aligned}
& =2 \tan ^{-1}\left[\operatorname{cosec} \tan ^{-1} x-\tan \cot ^{-1} x\right] \\
= & 2 \tan ^{-1}\left[\operatorname{cosec}\left\{\operatorname{cosec}^{-1} \frac{\sqrt{1+x^2}}{x}\right\}\right. \left.-\tan \left\{\tan ^{-1}\left(\frac{1}{x}\right)\right\}\right] \\
= & 2 \tan ^{-1}\left[\frac{\sqrt{1+x^2}}{x}-\frac{1}{x}\right]=2 \tan ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right] \\
= & 2 \tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right] {[putting \quad x=\tan \theta]} \\
= & \tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] \\
= & 2 \tan ^{-1}\left[\frac{2 \sin \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}\right] \\
= & 2 \tan ^{-1} \tan \frac{\theta}{2}=2 \cdot \frac{\theta}{2}=\theta=\tan ^{-1} x
\end{aligned}$
Hence, the correct answer is option: 'C'.