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Question: \[\sin^{- 1}\frac{12}{13} + \cos^{- 1}\frac{4}{5} + \tan^{- 1}\frac{63}{16} =\]...

sin⁡−11213+cos⁡−145+tan⁡−16316=\sin^{- 1}\frac{12}{13} + \cos^{- 1}\frac{4}{5} + \tan^{- 1}\frac{63}{16} =

A

0

B

π2\frac{\pi}{2}

C

π\pi

D

3π2\frac{3\pi}{2}

Answer

π\pi

Explanation

Solution

tan⁡−1125+tan⁡−134+tan⁡−16316\tan^{- 1}\frac{12}{5} + \tan^{- 1}\frac{3}{4} + \tan^{- 1}\frac{63}{16}

=π+tan⁡−148+1520−36+tan⁡−16316\pi + \tan^{- 1}\frac{48 + 15}{20 - 36} + \tan^{- 1}\frac{63}{16} (xy>1xy > 1)

= π−tan⁡−16316+tan⁡−16316=π\pi - \tan^{- 1}\frac{63}{16} + \tan^{- 1}\frac{63}{16} = \pi