Question

Prove: \(tan^{-1} \frac {63}{16} = sin^{-1} \frac {5}{13} + cos^{-1}\frac 35\)

Answer 1

Let sin^-1\(\frac {5}{13}\) = x. then sin x=\(\frac {5}{13}\) \(\implies\)cosx=\(\frac {12}{13}\)
tan x = \(\frac {5}{12}\) \(\implies\)x = tan^-1\(\frac {5}{12}\)
sin^-1\(\frac {5}{13}\) = tan^-1\(\frac {5}{12}\) …..…… (1) 
Let cos^-1\(\frac {3}{5}\) = y. then cos y=\(\frac {3}{5}\).\(\implies\)sin y = \(\frac 45\). 
tan y=\(\frac 43\)\(\implies\)y = tan^-1\(\frac 43\). 
therefore cos^-1\(\frac {3}{5}\) = tan^-1\(\frac {4}{3}\) ……….. (2)     Using (1) and (2), 
we have:
 R.H.S = sin^-1\(\frac {5}{13}\)+cos^-1\(\frac {3}{5}\)
           = tan^-1\(\frac {5}{12}\) + tan^-1\(\frac {4}{3}\) 
           = tan^-1\(\frac {\frac {5}{12}+ \frac 43}{1-\frac {5}{12}. \frac 43 }\)    [tan^-1 x+tan^-1 y]
           = tan^-1 \(\frac {x+y}{1-xy}\) 
           = tan^-1\(\frac {63}{16}\)
           = L.H.S