Question

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

• A. $0$
• B. ${\cot }^{-1}26$
• C. $\pi$
• D. $Noneofthese$

Answer 1

Answer: (C)

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