A) \[\cot \left( \frac{y-x}{2} \right)\]
B) \[\cot \left( \frac{x+y}{4} \right)\]
C) \[\cot \left( \frac{y-x}{4} \right)\]
D) \[\cot \left( \frac{y-x}{4} \right)\]
Correct Answer: D
Solution :
Given, \[\cos \,x=3\,\cos \,y\] or \[\frac{3}{1}=\frac{\cos \,x}{\cos \,y}\] Applying componendo------------------------------------------------------------------------- and dividendo, we get \[\frac{3+1}{3-1}=\frac{\cos \,x+\,\cos \,y}{\cos \,x-\,\cos y}\] \[\Rightarrow \] \[2=\frac{2\,\cos \left( \frac{x+y}{2} \right)\cos \left( \frac{x-y}{2} \right)}{2\sin \left( \frac{x+y}{2} \right)\sin \left( \frac{y-x}{2} \right)}\] \[\Rightarrow \] \[2=\cot \left( \frac{x+y}{2} \right)\cot \left( \frac{y-x}{2} \right)\] \[\Rightarrow \] \[2\tan \left( \frac{y-x}{2} \right)=\cot \left( \frac{x+y}{2} \right)\]
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