Q78 / 155 · WBJEE 2018 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'C'.

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