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. $\frac{m}{m+1}$

Answer 1

Answer: (C)

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

Given $\cos A=m\cos B$
and $\cot \left(\frac{A+B}{2}\right)=\lambda tan\left(\frac{B-A}{2}\right)$
$\lambda =\frac{\cot \left(\frac{A+B}{2}\right)}{\tan \left(\frac{B-A}{2}\right)}$
$=\frac{\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{B-A}{2}\right)}{\sin \left(\frac{A+B}{2}\right)sin\left(\frac{B-A}{2}\right)}$
$=\frac{\cos \left(B\right)+\cos \left(A\right)}{\cos A-\cos B}$
$\lambda =\frac{\frac{\cos A}{\cos B}+1}{\frac{\cos A}{\cos B}-1}$ (Dividing by cos B )
$\lambda =\frac{m+1}{m-1}$ $[m=\frac{\cos A}{\cos B}]$