If tan alpha = n tan beta and sin alpha = m s | vedclass

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(D) Given: $\sin \alpha = m \sin \beta \implies m = \frac{\sin \alpha}{\sin \beta} \implies m^2 - 1 = \frac{\sin^2 \alpha - \sin^2 \beta}{\sin^2 \beta

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$\cos^{2} \alpha$

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(D) Given: $\sin \alpha = m \sin \beta \implies m = \frac{\sin \alpha}{\sin \beta} \implies m^2 - 1 = \frac{\sin^2 \alpha - \sin^2 \beta}{\sin^2 \beta}$.Also,$	an \alpha = n 	an \beta \implies n = \frac{	an \alpha}{	an \beta} \implies n^2 - 1 = \frac{	an^2 \alpha - 	an^2 \beta}{	an^2 \beta}$.Now,$n^2 - 1 = \frac{\frac{\sin^2 \alpha}{\cos^2 \alpha} - \frac{\sin^2 \beta}{\cos^2 \beta}}{\frac{\sin^2 \beta}{\cos^2 \beta}} = \frac{\sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta}{\cos^2 \alpha \cos^2 \beta} \cdot \frac{\cos^2 \beta}{\sin^2 \beta} = \frac{\sin^2 \alpha (1 - \sin^2 \beta) - (1 - \sin^2 \alpha) \sin^2 \beta}{\cos^2 \alpha \sin^2 \beta}$.$= \frac{\sin^2 \alpha - \sin^2 \alpha \sin^2 \beta - \sin^2 \beta + \sin^2 \alpha \sin^2 \beta}{\cos^2 \alpha \sin^2 \beta} = \frac{\sin^2 \alpha - \sin^2 \beta}{\cos^2 \alpha \sin^2 \beta}$.Therefore,$\frac{m^2 - 1}{n^2 - 1} = \frac{\sin^2 \alpha - \sin^2 \beta}{\sin^2 \beta} \cdot \frac{\cos^2 \alpha \sin^2 \beta}{\sin^2 \alpha - \sin^2 \beta} = \cos^2 \alpha$.

If $\tan \alpha = n \tan \beta$ and $\sin \alpha = m \sin \beta,$ then $\frac{m^{2}-1}{n^{2}-1} =$

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