Question

If in $△ABC,a\tan A+b\tan B=(a+b)\tan \frac{1}{2}(A+B),$ prove that A= B.

Answer 1

We have $a[\tan A-\tan \frac{1}{2}(A+B\left)\right]$

$=b\left[\tan \frac{1}{2}\right(A+B)-\tan B]$ .....(1)

or $a\left[\frac{\sin A\cos \frac{1}{2}(A+B)-\cos A\sin \frac{1}{2}(A+B)}{\cos A\cos \frac{1}{2}(A+B)}\right]$

$=b\left[\frac{sin\frac{1}{2}(A+B)CosB-\cos \frac{1}{2}(A+B)\sin B}{\cos \frac{1}{2}(A+B)\cos B}\right]$

or $\frac{k\sin A\sin \frac{1}{2}(A-B)}{\cos A}=\frac{k\sin B\sin \frac{1}{2}(A-B)}{cosB}$

or $\sin \frac{1}{2}(A-B)(\frac{\sin A}{\cos A}-\frac{\sin A}{\cos B})=0$

or $\sin \frac{1}{2}(A-B)(\tan A-\tan B)=0$.

From this equation, we conclude that A=B.