Question

If tan A = n tan B and sin A = m sin B, prove that cos²A = $\frac{(m^2-1)}{n^2-1}.$

Answer 1

$$\begin{gathered} \begin{array}{l}\text{We have}\tan A=n\tan B\\ \end{array} \\ \text{=>}\cot B=\frac{n}{\tan A}...\left(\text{i}\right) \\ \text{Again,}\sin A=m\sin B \\ =>\cos \text{ec}B=\frac{m}{\sin A}...\left(\text{ii}\right) \\ \mathrm{S}\text{quaring}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{and subtracting}\left(\text{ii}\right)\text{from}\left(\text{i}\right),\mathrm{we}\mathrm{get} \\ \frac{m^2}{{\sin }^{2}A}-\frac{n^2}{{\tan }^{2}A}=\cos e{c}^{2}B-{\cot }^{2}B \\ =>\frac{m^2}{{\sin }^{2}A}-\frac{n^2\cos }{{\sin }^{2}A}=1 \\ =>m^2-n^2{\cos }^{2}A={\sin }^{2}A \\ =>m^2-n^2{\cos }^{2}A=1-{\cos }^{2}A \\ =>n^2{\cos }^{2}A-{\cos }^{2}A=m^2-1 \\ =>{\cos }^{2}A(n^2-1)=(m^2-1) \\ =>{\cos }^{2}A=\frac{(m^2-1)}{(n^2-1)} \\ \text{∴}{\cos }^{2}A=\frac{(m^2-1)}{(n^2-1)} \end{gathered}$$