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Mathematics Question on Inverse Trigonometric Functions

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

A

00

B

cot⁡−1 26\cot^{-1} \, 26

C

π\pi

D

None of theseNone\, of \,these

Answer

π\pi

Explanation

Solution

We have, cot⁡−1(21)+cot⁡−1(13)+cot⁡−1(−8)\cot^{-1} (21) + \cot^{-1} (13) + \cot^{-1} (-8)
=tan⁡−1(112)+tan⁡−1(113)+cot⁡−1(−8)= \tan^{-1}\left(\frac{1}{12}\right) +\tan^{-1}\left(\frac{1}{13}\right)+\cot^{-1}\left(-8\right)
=tan⁡−1(112+1131−121×113)+cot⁡−1(−8)= \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(34272)+(π−cot⁡−18)= \tan^{-1}\left(\frac{34}{272}\right) +\left(\pi - \cot^{-1} 8\right)
=tan⁡−1(18)+(−cot⁡−18+π)= \tan^{-1}\left(\frac{1}{8}\right) +\left( - \cot^{-1} 8 + \pi\right)
=tan⁡−1(18)−tan⁡−1(18)+π=π= \tan^{-1}\left(\frac{1}{8}\right) - \tan^{-1}\left(\frac{1}{8}\right) + \pi = \pi