Question

If $\cos A=m\cos B$, and $\cot \frac{A+B}{2}=\lambda \tan \frac{B-A}{2}$, then $\lambda$ is

• A. $\frac{m}{m-1}$
• B. $\frac{m+1}{m}$
• C. $\frac{m+1}{m-1}$
• D. None of these

Answer 1

The correct option is C $\frac{m+1}{m-1}$

We have, $\cos A=m\cos B$

$\Rightarrow \frac{\cos A}{\cos B}=\frac{m}{1}$

On applying componendo and dividendo rule, we get

$\Rightarrow \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}$

$\Rightarrow \frac{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{B-A}{2}\right)}{2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{B-A}{2}\right)}=\frac{m+1}{m-1}$

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

But, $\cot \frac{A+B}{2}=\lambda \tan \left(\frac{B-A}{2}\right)$

$\Rightarrow \lambda =\frac{m+1}{m-1}$.