MCQ
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
    $\frac{m+2}{m-2}$
  • $\frac{m+1}{m-1}$

Answer

Correct option: D.
$\frac{m+1}{m-1}$
Given:
$\cos A = m \cos B $
$\Rightarrow \frac{\cos A}{\cos B}=\frac{m}{1}$
$\Rightarrow \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}$
$\Rightarrow \frac{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}=\frac{m+1}{m-1}$
$\left[\because \cos A+\cos B =2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right) \text { and } \cos A -\cos B =-2 \sin \left(\frac{A+B}{2}\right) \cos \right.$
$\left.\left(\frac{A-B}{2}\right)\right]$
$\Rightarrow \frac{\cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{-\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}=\frac{m+1}{m-1}$
$\Rightarrow-\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)=\frac{m+1}{m-1}$
$\Rightarrow \cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)=\frac{1+m}{1-m}$

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