Q67 / 180 · COMEDK 2015 · Mathematics

$\cot ^{-1}(21)+\cot ^{-1}(13)+\cot ^{-1}(-8)=$

  • A
    0
  • B
    $\cot ^{-1} 26$
  • Correct answer
    C
    $\pi$
  • D
    None of these

Solution

We have, $\cot ^{-1}(21)+\cot ^{-1}(13)+\cot ^{-1}(-8)$ $=\tan ^{-1}\left(\frac{1}{21}\right)+\tan ^{-1}\left(\frac{1}{13}\right)+\cot ^{-1}(-8)$ $=\tan ^{-1}\left(\frac{\frac{1}{21}+\frac{1}{13}}{1-\frac{1}{21} \times \frac{1}{13}}\right)+\cot ^{-1}(-8)$ $=\tan ^{-1}\left(\frac{34}{272}\right)+\left(\pi-\cot ^{-1} 8\right)$ $\quad \quad \left[\because \cot ^{-1}(-x)=\pi-\cot ^{-1} x\right]$ $=\tan ^{-1}\left(\frac{1}{8}\right)-\cot ^{-1} 8+\pi$ $=\tan ^{-1}\left(\frac{1}{8}\right)-\tan ^{-1}\left(\frac{1}{8}\right)+\pi=\pi$

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

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