Prove that tan^-1(63/10) =Sin^-1 (5/13) + Cos^-1 (3/5)

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Show that $s i n^{- 1} \frac{5}{13} + c o s^{- 1} \frac{3}{5} = t a n^{- 1} \frac{63}{16} .$

{sin}^{- 1} \frac{5}{13} + {cos}^{- 1} \frac{3}{5} = {tan}^{- 1} \frac{63}{16}

{cos}^{- 1} \frac{63}{65} + 2 {tan}^{- 1} \frac{1}{5} = {sin}^{- 1} \frac{3}{5}

sin^{- 1} \frac{3}{5} + cos^{- 1} \frac{12}{13} = sin^{- 1} (\frac{56}{65})

$t a n^{- 1} \frac{63}{16} = s i n^{- 1} \frac{5}{13} + c o s^{- 1} \frac{3}{5}$

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