Q.If \( \sin x = m \sin y \) and \( \tan x = n \tan y \), then prove that \[ \cos^2 x = \cfrac{m^2 - 1}{n^2 - 1} \]

Given: \[ \sin x = m \sin y \quad \text{and} \quad \tan x = n \tan y \] Then, \[ \sin^2 x = m^2 \sin^2 y \Rightarrow 1 - \cos^2 x = m^2 (1 - \cos^2 y) \Rightarrow 1 - \cos^2 x = m^2 - m^2 \cos^2 y \Rightarrow m^2 \cos^2 y = m^2 - 1 + \cos^2 x \Rightarrow \cos^2 y = \cfrac{m^2 - 1 + \cos^2 x}{m^2} \quad \text{—(i)} \] Also, \[ \tan x = n \tan y \Rightarrow \cfrac{\sin x}{\cos x} = n \cfrac{\sin y}{\cos y} \Rightarrow \cfrac{m \sin y}{\cos x} = n \cfrac{\sin y}{\cos y} \quad \text{[using } \sin x = m \sin y\text{]} \Rightarrow \cfrac{m}{\cos x} = \cfrac{n}{\cos y} \Rightarrow \cos y = \cfrac{n \cos x}{m} \Rightarrow \cos^2 y = \cfrac{n^2 \cos^2 x}{m^2} \quad \text{—(ii)} \] From equations (i) and (ii), \[ \cfrac{m^2 - 1 + \cos^2 x}{m^2} = \cfrac{n^2 \cos^2 x}{m^2} \Rightarrow m^2 - 1 + \cos^2 x = n^2 \cos^2 x \Rightarrow n^2 \cos^2 x - \cos^2 x = m^2 - 1 \Rightarrow \cos^2 x (n^2 - 1) = m^2 - 1 \Rightarrow \cos^2 x = \cfrac{m^2 - 1}{n^2 - 1} \quad \text{[Proved]} \]
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