If tan A = n tan B and sin A = m sin B , then | vedclass

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(A) Given: $	an A = n 	an B$ and $\sin A = m \sin B$.From the first equation: $\frac{\sin A}{\cos A} = n \frac{\sin B}{\cos B} \Rightarrow \frac{\si

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$\frac{m^{2}-1}{n^{2}-1}$

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(A) Given: $	an A = n 	an B$ and $\sin A = m \sin B$.From the first equation: $\frac{\sin A}{\cos A} = n \frac{\sin B}{\cos B} \Rightarrow \frac{\sin A}{\sin B} = n \frac{\cos A}{\cos B}$.Since $\sin A = m \sin B$,we have $\frac{\sin A}{\sin B} = m$.Equating the two expressions for $\frac{\sin A}{\sin B}$,we get $m = n \frac{\cos A}{\cos B}$,which implies $\cos B = \frac{n}{m} \cos A$.Also,from $\sin A = m \sin B$,we get $\sin B = \frac{1}{m} \sin A$.Using the identity $\sin^2 B + \cos^2 B = 1$:$(\frac{1}{m} \sin A)^2 + (\frac{n}{m} \cos A)^2 = 1$.$\frac{1}{m^2} \sin^2 A + \frac{n^2}{m^2} \cos^2 A = 1$.$\sin^2 A + n^2 \cos^2 A = m^2$.Substituting $\sin^2 A = 1 - \cos^2 A$:$(1 - \cos^2 A) + n^2 \cos^2 A = m^2$.$(n^2 - 1) \cos^2 A = m^2 - 1$.Therefore,$\cos^2 A = \frac{m^2 - 1}{n^2 - 1}$.

If $\tan A = n \tan B$ and $\sin A = m \sin B ,$ then the value of $\cos ^{2} A$ is

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