A) \[A=B\]
B) \[~A=-B\]
C) \[A=2B\]
D) \[B=2A.\]
Correct Answer: A
Solution :
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) \[\Rightarrow \]\[\,\tan A\sin \left( \frac{A-B}{2} \right)=\tan B\sin \left( \frac{A-B}{2} \right)\] \[\Rightarrow \]\[\sin \left( \frac{A-B}{2} \right)(tanA-\tan B)=0\] \[\Rightarrow \]\[A=B\]
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