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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