If $\cos A=m \cos B$, then $\cot \frac{A+B}{2} \cot \frac{B-A}{2}=$
(a) $\frac{m-1}{m+1}$
(b) $\frac{m+2}{m-2}$
(c) $\frac{m+1}{m-1}$
(d) None of these
(c) $\frac{m+1}{m-1}$
Given:
$\cos A=m \cos B$
$\Rightarrow \frac{\cos A}{\cos B}=\frac{m}{1}$
$\Rightarrow \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}$
$\Rightarrow \frac{2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{-2 \sin \left(\frac{B+A}{2}\right) \sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1}$
$\left[\because \cos A+\cos B=2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right.$ and $\left.\cos A-\cos B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{B-A}{2}\right)\right]$
$\Rightarrow \frac{\cos \left(\frac{B-A}{2}\right) \cos \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1}$
$\Rightarrow \cot \left(\frac{A+B}{2}\right) \cot \left(\frac{B-A}{2}\right)=\frac{m+1}{m-1}$
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