If tanA = n tanB and sinA = m sinB, then the value of cos²A = ?

Solution (By Examveda Team)

$$\eqalign{ & {\text{sin A}} = m{\text{ sin B}} \cr & {\text{si}}{{\text{n}}^2}{\text{A}} = {m^2}{\text{si}}{{\text{n}}^2}{\text{B }}......{\text{(i)}} \cr & {\text{Now, ta}}{{\text{n}}^2}{\text{A}} = {n^2}{\text{ta}}{{\text{n}}^2}{\text{B}} \cr & \frac{{{{\sin }^2}{\text{A}}}}{{{\text{co}}{{\text{s}}^2}{\text{A}}}} = {n^2}\frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & {\text{from equation (i)}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{{\sin }^2}{\text{B}}}}{{{\text{co}}{{\text{s}}^2}{\text{B}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{\frac{{\left( {1 - {\text{co}}{{\text{s}}^2}{\text{A}}} \right)}}{{{m^2}}}}}{{1 - \frac{{{{\sin }^2}{\text{A}}}}{{{m^2}}}}} \cr & \Rightarrow \frac{{1 - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}}}} = \frac{{{\text{1}} - {\text{co}}{{\text{s}}^2}{\text{A}}}}{{{m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}}}} \cr & \Rightarrow {m^2} - 1 + {\text{co}}{{\text{s}}^2}{\text{A}} = {{\text{n}}^2}{\text{co}}{{\text{s}}^2}{\text{A}} \cr & \Rightarrow {m^2} - 1 = {\text{co}}{{\text{s}}^2}\theta \left( {{n^2} - 1} \right) \cr & \Rightarrow {\text{co}}{{\text{s}}^2}{\text{A}} = \frac{{{m^2} - 1}}{{{n^2} - 1}} \cr} $$