Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(sin^{-1}{56\over 65}\)

Solution :

We have, L.H.S. = \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(tan^{-1}{5\over 12}\) + \(tan^{-1}{3\over 4}\)

\(\because\) [ \(cos^{-1}{12\over 13}\) = \(tan^{-1}{5\over 12}\) & \(sin^{-1}{3\over 5}\) = \(tan^{-1}{3\over 4}\) ]

L.H.S. = \(tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})\) = \(tan^{-1}{56\over 33}\)

R.H.S. = \(sin^{-1}{56\over 65}\) = \(tan^{-1}{56\over 33}\)

L.H.S = R.H.S.  Hence Proved.


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