If $${\sin ^{ - 1}}1 + {\sin ^{ - 1}}\frac{4}{5} = {\sin ^{ - 1}}x,$$      then what is $$x$$ equal to ?

Solution :
$$\eqalign{ & {\text{Let,}}\,\,{\sin ^{ - 1}}\left( 1 \right) + {\sin ^{ - 1}}\left( {\frac{4}{5}} \right) = {\sin ^{ - 1}}x \cr & {\text{Let,}}\,\,{\sin ^{ - 1}}\left( 1 \right) = \theta \cr & \Rightarrow \sin \theta = 1 \cr & \Rightarrow \cos \theta = 0\,\,{\text{and }}{\sin ^{ - 1}}\left( {\frac{4}{5}} \right) = \phi \cr & \Rightarrow \sin \phi = \left( {\frac{4}{5}} \right) \cr & \Rightarrow \cos \phi = \sqrt {1 - \frac{{16}}{{25}}} = \sqrt {\frac{9}{{25}}} = \frac{3}{5} \cr & \therefore {\sin ^{ - 1}}x = \theta + \phi \cr & \Rightarrow x = \sin \left( {\theta + \phi } \right) = \sin \theta \cos \phi + \cos \theta \sin \phi \cr & = 1 \times \frac{3}{5} + 0 \times \frac{4}{5} \cr & \Rightarrow x = \frac{3}{5} \cr} $$