Question

If cos A = m cos B, then write the value of $\cot \frac{A+B}{2}\cot \frac{A-B}{2}$.

Answer 1

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

$\Rightarrow \mathrm{c}\mathrm{ot}\left(\frac{A+B}{2}\right)\cot \left(\frac{A-B}{2}\right)=\frac{1+m}{1-m}$