Question

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

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

Answer 1

Answer: (C)

We have, $\cot^{-1} (21) + \cot^{-1} (13) + \cot^{-1} (-8) $
$= \tan^{-1}\left(\frac{1}{12}\right) +\tan^{-1}\left(\frac{1}{13}\right)+\cot^{-1}\left(-8\right) $
$= \tan^{-1}\left(\frac{\frac{1}{12}+\frac{1}{13}}{1-\frac{1}{21}\times\frac{1}{13}}\right) +\cot^{-1}\left(-8\right) $
$= \tan^{-1}\left(\frac{34}{272}\right) +\left(\pi - \cot^{-1} 8\right) $
$= \tan^{-1}\left(\frac{1}{8}\right) +\left( - \cot^{-1} 8 + \pi\right) $
$ = \tan^{-1}\left(\frac{1}{8}\right) - \tan^{-1}\left(\frac{1}{8}\right) + \pi = \pi $