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$\frac{\cos x}{\cos (x-2 y)}=\lambda \Rightarrow \tan (x-y) \tan y$ is equal to
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Verified Answer
The correct answer is:
$\frac{1-\lambda}{1+\lambda}$
$\begin{aligned}
tan(x-y) \tan y=\frac{\sin (x-y) \sin y}{\cos (x-y) \cos y} \times \frac{2}{2} \\
=\frac{\cos (x-2 y)-\cos (x)}{\cos (x-2 y)+\cos (x)} \\
=\frac{1-\frac{\cos x}{\cos (x-2 y)}}{1+\frac{\cos (x)}{\cos (x-2 y)}} \\
=\frac{1-\lambda}{1+\lambda} \\
&\left(\text { Given, } \lambda=\frac{\cos x}{\cos (x-2 y)}\right)
\end{aligned}$
tan(x-y) \tan y=\frac{\sin (x-y) \sin y}{\cos (x-y) \cos y} \times \frac{2}{2} \\
=\frac{\cos (x-2 y)-\cos (x)}{\cos (x-2 y)+\cos (x)} \\
=\frac{1-\frac{\cos x}{\cos (x-2 y)}}{1+\frac{\cos (x)}{\cos (x-2 y)}} \\
=\frac{1-\lambda}{1+\lambda} \\
&\left(\text { Given, } \lambda=\frac{\cos x}{\cos (x-2 y)}\right)
\end{aligned}$
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