Question

For any triangle $ABC$, prove that
$\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}$

Answer 1

We have,

$\frac{a+b}{c}$

$=\frac{2R(\sin A+\sin B)}{2R\sin C}$ [ Using sine rule of triangle]

$=\frac{\sin A+\sin B}{\sin C}$

$=\frac{2.\sin \frac{A+B}{2}.\cos \frac{A-B}{2}}{2.\sin \frac{C}{2}.\cos \frac{C}{2}}$

$=\frac{\sin (\frac{\pi }{2}-\frac{C}{2}).\cos \frac{A-B}{2}}{\sin \frac{C}{2}.\cos \frac{C}{2}}$ [ Since $A+B+C={180}^o$]

$=\frac{\cos \frac{C}{2}.\cos \frac{A-B}{2}}{\sin \frac{C}{2}.\cos \frac{C}{2}}$

$=\frac{\cos \frac{A-B}{2}}{\sin \frac{C}{2}}$