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 = \(\frac {8}{17}\)\(\implies\)cos 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