Question

If $0\le A\le \frac{\pi }{4},$ then ${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}(\cot A)+{\tan }^{-1}({\cot }^{3}A)$ is equal to

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

We have, $0\le A\le \frac{\pi }{4}$
${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}(\cot A)+{\tan }^{-1}\left({\cot }^{3}A\right)$
$={\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\frac{\cot A+{\cot }^{3}A}{1-{\cot }^{4}A}\right)$
$={\tan }^{-1}\left(\frac{1}{2}\cdot \frac{2\tan A}{1-{\tan }^{2}A}\right)+{\tan }^{-1}\left(\frac{\tan A}{{\tan }^{2}A-1}\right)$
$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)-{\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)$
$=0$