A) \[\frac{1+k}{1-k}\]
B) \[\frac{1-k}{1+k}\]
C) \[\frac{2k}{k+1}\]
D) \[\frac{k-1}{2k+1}\]
Correct Answer: B
Solution :
\[\cos x=k\cos \,\,(x-2y)\frac{1}{k}=\frac{\cos \,\,(x-2y)}{\cos x}\] Apply componendo and dividendo theorem \[\frac{1-k}{1+k}=\frac{\cos \,\,(x-2y)-\cos x}{\cos \,\,(x-2y)+\cos x}\] \[=\frac{2\sin \,\,(x-y)\sin y}{2\cos \,\,(x-y)\cos y}=\tan \,\,(x-y)\tan y\]
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