A) \[\frac{m}{m-1}\]
B) \[\frac{m+1}{m}\]
C) \[\frac{m+1}{m-1}\]
D) None of these
Correct Answer: C
Solution :
We have,\[\cos A=m\cos B\] \[\Rightarrow \] \[\frac{\cos A}{\cos B}=\frac{m}{1}\] On applying componendo and dividendo rule, we get \[\Rightarrow \] \[\frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}\] \[\Rightarrow \] \[\frac{2\cos \left( \frac{A+B}{2} \right)\cos \left( \frac{B-A}{2} \right)}{2\sin \left( \frac{A+B}{2} \right)\sin \left( \frac{B-A}{2} \right)}=\frac{m+1}{m-1}\] \[\Rightarrow \] \[\cot \left( \frac{A+B}{2} \right)=\left( \frac{m+1}{m-1} \right)\tan \left( \frac{B-A}{2} \right)\] But, \[\cot \frac{A+B}{2}=\lambda \tan \left( \frac{B-A}{2} \right)\] (given) \[\Rightarrow \] \[\lambda =\frac{m+1}{m-1}\]
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