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

Answer: Option C

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