Question

In a triangle $ABC$, prove that
$(a+b)\cos \frac{A+B}{2}=c\cos \frac{A-B}{2}$

Answer 1

We have to prove that

$\left(a+b\right)\cos \frac{A+B}{2}=c\cos \frac{A-B}{2}$

Now,

$L.H.S=\left(a+b\right)\cos \frac{A+B}{2}$

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

$=k\left[\sin A+\sin B\right]\sin \frac{c}{2}$

$=k\left[2\sin \frac{A+B}{2}.\cos \frac{A-B}{2}\right]\sin \frac{c}{2}$

$=k\left[2\cos \frac{c}{2}\sin \frac{c}{2}\cos \frac{A-B}{2}\right]$

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

$=\frac{c}{\sin c}\sin c\cos \frac{A-B}{2}$ [cancle $sinc$ ]

$=c\cos \frac{A-B}{2}=R.H.S$

Hence

$L.H.S=R.H.S$

Proved.