Question

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

Answer 1

(c)$\frac{m+1}{m-1}$


$$\begin{gathered} \text{Given}: \\ \cos A=m\cos B \\ \Rightarrow \frac{\cos A}{\cos B}=\frac{m}{1} \\ \Rightarrow \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1} \\ \Rightarrow \frac{2\cos \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)}{-2\sin \left(\frac{B+A}{2}\right)\sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1}\left[\because \cos A+\cos B=2\cos \left(\frac{A-B}{2}\right)\cos \left({\displaystyle \frac{A+B}{2}}\right)\mathrm{and}\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\cos \left({\displaystyle \frac{B-A}{2}}\right)\right] \\ \Rightarrow \frac{\cos \left(\frac{B-A}{2}\right)\cos \left(\frac{A+B}{2}\right)}{\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1} \\ \Rightarrow \mathrm{c}\mathrm{ot}\left(\frac{A+B}{2}\right)\cot \left(\frac{B-A}{2}\right)=\frac{m+1}{m-1} \end{gathered}$$