If A = m B, then A+B 2 B-A 2 =

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
$\frac{m+2}{m-2}$
D
$\frac{m+1}{m-1}$

✓

Answer

(d) $\frac{m+1}{m-1}$
Explanation: Given:
cos A = m cos B
$\begin{array}{l}\Rightarrow \frac{\cos A}{\cos B}=\frac{m}{1} \\ \Rightarrow \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1} \\ \Rightarrow \frac{2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}{-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}=\frac{m+1}{m-1}\left[\because \cos A+\cos B =2 \cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right) \text { and } \cos A -\cos B =-2 \sin \left(\frac{A+B}{2}\right) \cos \right. \\ \left.\left(\frac{A-B}{2}\right)\right] \\ \Rightarrow \frac{\cos \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)}{-\sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}=\frac{m+1}{m-1} \\ \Rightarrow-\cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)=\frac{m+1}{m-1} \\ \Rightarrow \cot \left(\frac{A+B}{2}\right) \cot \left(\frac{A-B}{2}\right)=\frac{1+m}{1-m}\end{array}$

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