A) \[\cot \frac{A+B}{2}=\frac{m+1}{m-1}\tan \frac{B-A}{2}\]
B) \[\tan \frac{A+B}{2}=\frac{m+1}{m-1}\cot \frac{B-A}{2}\]
C) \[\cot \frac{A+B}{2}=\frac{m+1}{m-1}\tan \frac{A-B}{2}\]
D) None of these
Correct Answer: A
Solution :
Given that \[\cos A=m\,\,\cos B\,\Rightarrow \,\,\frac{m}{1}=\frac{\cos A}{\cos B}\] \[\Rightarrow \,\,\frac{m+1}{m-1}=\frac{\cos A+\cos B}{\cos A-\cos B}=\frac{2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{B-A}{2} \right)}{2\sin \left( \frac{A+B}{2} \right)\sin \left( \frac{B-A}{2} \right)}\] \[=\cot \,\left( \frac{A+B}{2} \right)\,\cot \,\left( \frac{B-A}{2} \right)\] Hence, \[\cot \,\left( \frac{A+B}{2} \right)=\frac{m+1}{m-1}\tan \frac{B-A}{2}\].
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