0 < θ < 90°, tanθ + sinθ =m and tanθ - sinθ = n, where m ≠ n, then the value of m² - n² is?

Solution (By Examveda Team)

$$\eqalign{ & \tan \theta + \sin \theta = m \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta + 2{\text{ tan}}\theta .\sin \theta = {m^2}\,....(i) \cr & {\text{tan}}\theta - \sin \theta = n \cr & {\text{Squaring both sides}} \cr & {\tan ^2}\theta + {\sin ^2}\theta - 2{\text{ tan}}\theta .\sin \theta = {n^2}\,....(ii) \cr & {\text{Substract from (i) and (ii)}} \cr & {m^2} - {n^2} = {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta - {\text{ta}}{{\text{n}}^2}\theta - {\sin ^2}\theta + 2{\text{tan}}\theta \sin \theta \cr & {m^2} - {n^2} = 4{\text{tan}}\theta \sin \theta \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta {{\sin }^2}\theta } \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \left( {1 - {\text{co}}{{\text{s}}^2}\theta } \right)} \cr & = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta - {{\sin }^2}\theta } \cr & = 4\sqrt {mn} \cr} $$