• # question_answer If in a$\text{ABC,}$ $a\tan A+b\,\tan \,B=(a+b)\tan \left( \frac{A+B}{2} \right),$then A) $A=B$            B)        $~A=-B$            C)        $A=2B$           D)         $B=2A.$

We have, $a\tan \,A+b\tan B=(a+b)\tan \left( \frac{A+B}{2} \right)$ $\Rightarrow$$a\left[ \tan A-\tan \left( \frac{A+B}{2} \right) \right]$                 $=b\left[ \tan \left( \frac{A+B}{2} \right)-\tan B \right]$                 $a\left[ \frac{\sin A\cos \left( \frac{A+B}{2} \right)-\cos A\sin \left( \frac{A+B}{2} \right)}{\cos A.\cos \left( \frac{A+B}{2} \right)} \right]$ $=b\left[ \frac{\cos B\sin \left( \frac{A+B}{2} \right)-\sin B\cos \left( \frac{A+B}{2} \right)}{\cos B.\cos \left( \frac{A+B}{2} \right)} \right]$ $\Rightarrow$$\frac{a\sin \left\{ A-\left( \frac{A+B}{2} \right) \right\}}{\cos A}=\frac{b\sin \left\{ \left( \frac{A+B}{2} \right)-B \right\}}{\cos B}$ $\Rightarrow$$\frac{a\sin \left( \frac{A-B}{2} \right)}{\cos A}=\frac{b\sin \left( \frac{A-B}{2} \right)}{\cos B}$ $\Rightarrow$$\frac{k\sin A\sin \left( \frac{A-B}{2} \right)}{\cos A}=\frac{k\sin Bsin\left( \frac{A-B}{2} \right)}{\cos B}$                                 (using sine rule) $\Rightarrow$$\,\tan A\sin \left( \frac{A-B}{2} \right)=\tan B\sin \left( \frac{A-B}{2} \right)$ $\Rightarrow$$\sin \left( \frac{A-B}{2} \right)(tanA-\tan B)=0$ $\Rightarrow$$A=B$