$$\frac{{{\text{cos }}\alpha }}{{{\text{sin }}\beta }} = n$$   and $$\frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} = m,$$   then the value of $${\text{co}}{{\text{s}}^2}\beta $$   is?

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given,}} \cr & n = \frac{{{\text{cos }}\alpha }}{{{\text{sin }}\beta }}{\text{,}}\,\,\,\,\,\,{\text{m}} = \frac{{{\text{cos }}\alpha }}{{{\text{cos }}\beta }} \cr & \Rightarrow {\text{cos }}\alpha = n{\text{ sin }}\beta ,\,and\,\cos \alpha = m\,\cos \beta \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{ }}\alpha = {n^2}{\text{ si}}{{\text{n}}^2}{\text{ }}\beta .....(i) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}\alpha = {m^2}{\text{ co}}{{\text{s}}^2}{\text{ }}\beta .....(ii) \cr & equation{\text{ }}(i) = (ii) \cr & \Rightarrow {n^2}{\text{ si}}{{\text{n}}^2}\beta = {m^2}{\text{ co}}{{\text{s}}^2}\beta \cr & \Rightarrow {n^2}\left( {{\text{1}} - {\text{ co}}{{\text{s}}^2}\beta } \right) = {m^2}{\text{ co}}{{\text{s}}^2}\beta \cr & \Rightarrow {n^2} - {n^2}{\cos ^2}\beta = {m^2}{\cos ^2}\beta \cr & \Rightarrow {n^2} = {m^2}{\cos ^2}\beta + {n^2}{\cos ^2}\beta \cr & \Rightarrow {n^2} = {\cos ^2}\beta \left( {{m^2} + {n^2}} \right) \cr & \Rightarrow {\cos ^2}\beta = \frac{{{n^2}}}{{{m^2} + {n^2}}} \cr} $$