If frac{cos x}{cos (x-2y)} = lambda,then tan | vedclass

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(B) We have $	an (x-y) 	an y = \frac{\sin (x-y) \sin y}{\cos (x-y) \cos y}$.Multiplying numerator and denominator by $2$,we get $\frac{2 \sin (x-y)

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$\frac{1-\lambda}{1+\lambda}$

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(B) We have $	an (x-y) 	an y = \frac{\sin (x-y) \sin y}{\cos (x-y) \cos y}$.Multiplying numerator and denominator by $2$,we get $\frac{2 \sin (x-y) \sin y}{2 \cos (x-y) \cos y}$.Using the product-to-sum formulas $2 \sin A \sin B = \cos (A-B) - \cos (A+B)$ and $2 \cos A \cos B = \cos (A-B) + \cos (A+B)$:$	an (x-y) 	an y = \frac{\cos (x-2y) - \cos x}{\cos (x-2y) + \cos x}$.Dividing numerator and denominator by $\cos (x-2y)$:$= \frac{1 - \frac{\cos x}{\cos (x-2y)}}{1 + \frac{\cos x}{\cos (x-2y)}}$.Substituting $\lambda = \frac{\cos x}{\cos (x-2y)}$,we get $\frac{1-\lambda}{1+\lambda}$.

If $\frac{\cos x}{\cos (x-2y)} = \lambda$,then $\tan (x-y) \tan y$ is equal to

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