Question

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

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

Answer 1

The correct option is B $\pi$

${\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\cot A\right)+{\tan }^{-1}\left({\cot }^{3}A\right)$

$={\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\frac{\cot A+{\cot }^{3}A}{1-{\cot }^{4}A}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\tan 2A\right)+{\tan }^{-1}\left(\frac{\cot A(1+{\cot }^{2}A)}{(1-{\cot }^{2}A)(1+{\cot }^{2}A)}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\frac{2\tan A}{1-{\tan }^{2}A}\right)+{\tan }^{-1}\left(\frac{\tan A}{{\tan }^{2}A-1}\right)$

$={\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)+\pi -{\tan }^{-1}\left(\frac{\tan A}{1-{\tan }^{2}A}\right)$

$=\pi$