A) \[\frac{1+\lambda }{1-\lambda }\]
B) \[\frac{1-\lambda }{1+\lambda }\]
C) \[\frac{\lambda }{1+\lambda }\]
D) \[\frac{\lambda }{1-\lambda }\]
Correct Answer: B
Solution :
Now,\[\tan (x-y)\tan y\] \[=\frac{\sin (x-y)\sin y}{\cos (x-y)\cos y}\times \frac{2}{2}\] \[=\frac{\cos (x-2y)-\cos (x)}{\cos (x-2y)+\cos (x)}\] \[=\frac{1-\frac{\cos x}{\cos (x-2y)}}{1+\frac{\cos (x)}{\cos (x-2y)}}\] \[=\frac{1-\lambda }{1+\lambda }\] \[\left( Given,\,\,\lambda =\frac{\cos x}{\cos (x-2y)} \right)\]
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