Given that:$a\tan A+b\tan B=(a+b)·\tan \left(\frac{A+B}{2}\right)$

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

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

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

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

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

$\Rightarrow \sin \left(\frac{A-B}{2}\right)\left(\frac{a}{\cos A}-\frac{b}{\cos B}\right)=0$

As we know that$\frac{a}{\sin A}=\frac{b}{\sin B}=k$

$\Rightarrow \sin \left(\frac{A-B}{2}\right)\left(\frac{k\sin A}{\cos A}-\frac{k\sin B}{\cos B}\right)=0$

$\Rightarrow \sin \left(\frac{A-B}{2}\right)\left(\tan A-\tan B\right)=0$

$\Rightarrow \sin \left(\frac{A-B}{2}\right)=0 \mathrm{or} (\mathrm{tanA}-\mathrm{tanB})=0$

$\Rightarrow A=B$