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$$\impliesy = 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