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

Answer: (C)

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

solution:-

Given $cosA=mcosB$

$\Rightarrow m=\frac{cosA}{coB}$

we have to show find $cot\frac{A+B}{2}cot\frac{B-A}{2}$

Let $\frac{cot\frac{A+B}{2}}{tan\frac{B-A}{2}}=\frac{cot\frac{A+B}{2}}{-tan\frac{A-B}{2}}$

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

$=\frac{cosA+cosB}{cosA-cosB}$

$=\frac{\frac{cosA}{cosB}+1}{\frac{cosA}{cosB}-1}=\frac{m+1}{m-1}$