Question

If sin A = n sin B, then $\frac{n-1}{n+1}$ tan $\frac{A+B}{2}$ =

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We have sin A = n sin B $\Rightarrow$ $\frac{n}{1}$ = $\frac{sinA}{sinB}$

$\Rightarrow$ $\frac{n-1}{n+1}$ $\frac{sinA-sinB}{sinA+sinB}$ = $\frac{2cos\frac{A+B}{2}sin\frac{A-B}{2}}{2sin\frac{A+B}{2}cos\frac{A-B}{2}}$

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

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