A) \[A=B\]
B) \[B=C\]
C) \[C=A\]
D) \[A=B=C\]
Correct Answer: A
Solution :
Given that \[a\tan A+b\tan B=(a+b)\tan \left( \frac{A+B}{2} \right)\] \[\Rightarrow \] \[a\left[ \tan A-\tan \left( \frac{A+B}{2} \right) \right]\] \[=-b\left[ \tan B-\tan \left( \frac{A+B}{2} \right) \right]\] \[\Rightarrow \] \[a\frac{\left[ \begin{align} & \sin A\cos \left( \frac{A+B}{2} \right) \\ & -\cos A\sin \left( \frac{A+B}{2} \right) \\ \end{align} \right]}{\cos A\cos \left( \frac{A+B}{2} \right)}\] \[=-b\left[ \frac{\begin{align} & \sin B\cos \left( \frac{A+B}{2} \right) \\ & -\sin \left( \frac{A+B}{2} \right)\cos B \\ \end{align}}{\cos B\cos \left( \frac{A+B}{2} \right)} \right]\] \[\Rightarrow \]\[\frac{k\sin A\sin \left( \frac{A-B}{2} \right)}{\cos A}=\frac{k\sin B\sin \left( \frac{A-B}{2} \right)}{\cos B}\] \[\left[ \because \,\,\frac{a}{\sin A}=\frac{b}{\sin B}=k \right]\] \[\Rightarrow \] \[\sin \left( \frac{A-B}{2} \right)(\tan A-\tan B)=0\] \[\Rightarrow \] \[\tan A-\tan B=0\Rightarrow A=B\]
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