Question

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