Question

Prove: $sin^{-1} \frac {8}{17} + sin^{-1} \frac 35=tan^{-1} \frac {77}{36}$

Answer 1

Let sin-1$\frac {8}{17}$ = x. Then, sin x = \frac {8}{17}$$\impliescos x =$\sqrt {1-(\frac {8}{17})^2}$ = $\sqrt {\frac {225}{289}}$=$\frac {15}{17}$.
therefore tan x = $\frac {8}{15}$ $\implies$x = tan-1$\frac {8}{15}$
so sin-1$\frac {8}{17}$ = tan-1$\frac {8}{15}$ …...…. (1)
Now let sin-1$\frac {3}{5}$ = y Then sin y=$\frac {3}{5}$.$\implies$cos y=$\sqrt {1-(\frac {3}{5})^2}$ = $\sqrt {\frac {16}{25}}$ = $\frac {4}{5}$.
tan y = $\frac {3}{4}$, y= tan-1$\frac {3}{4}$
therefore sin-1$\frac {3}{5}$ = tan-1$\frac {3}{4}$ .……….. (2)
Now, we have:
LHS= sin-1$\frac {8}{17}$ + sin-1$\frac {3}{5}$
=tan-1 $\frac {8}{15}$ + tan-1$\frac {3}{4}$ [using(1) and (2)]
=tan-1$\frac {\frac {8}{15}+ \frac 3 4}{1-\frac {8}{15}. \frac 3 4 }$
=tan-1$\frac {77}{36}$
=RHS