Let sin-1 (\(\frac{4}{5}\)) = A
sin A = \(\frac{4}{5}\)
∴ cos A = \(\frac{3}{5}\)
Let sin-1 (\(\frac{12}{13}\)) = B
\(\frac{12}{13}\) = sin B
sin B = \(\frac{12}{13}\)
∴ cos B = \(\frac{5}{13}\)
Now cos(sin-1(\(\frac{4}{5}\)) + sin-1 (\(\frac{12}{13}\))) = cos (A + B)
= cos A cos B – sin A sin B