# 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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