If $$\frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a$$   and $$\frac{{{\text{sin }}\alpha }}{{{\text{sin }}\beta }} = b{\text{,}}$$   then the value of $${\sin ^2}\beta $$  in terms of a and b is?

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = a{\text{ }} \cr & \Rightarrow \cos {\text{ }}\alpha = a{\text{ }}\cos {\text{ }}\beta \cr & {\text{On squaring both sides}} \cr & {\cos ^2}\alpha = {a^2}{\cos ^2}\beta \cr & \Rightarrow 1 - {\sin ^2}\alpha = {a^2}\left( {1 - {{\sin }^2}\beta } \right)....(i) \cr & {\text{Again, }}\sin \alpha = {\text{ }}b\sin \beta \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\sin ^2}\alpha = {\text{ }}{b^2}{\sin ^2}\beta \cr & {\text{Put the value of }}{\sin ^2}\alpha {\text{ in equation (i)}} \cr & \Rightarrow {\text{1}} - {b^2}{\sin ^2}\beta = {a^2} - {a^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = {a^2}si{n^2}\beta - {b^2}si{n^2}\beta \cr & \Rightarrow {a^2} - 1 = si{n^2}\beta \left( {{a^2} - {b^2}} \right) \cr & \Rightarrow si{n^2}\beta = \frac{{{a^2} - 1}}{{{a^2} - {b^2}}} \cr} $$