If in triangle ABC,a tan A + b tan B = (a + b | vedclass

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(A) Given that: $a 	an A + b 	an B = (a + b) 	an \left(\frac{A+B}{2}ight)$ Rearranging the terms: $a \left[ 	an A - 	an \left(\frac{A+B}{2}ig

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$A = B$

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(A) Given that: $a 	an A + b 	an B = (a + b) 	an \left(\frac{A+B}{2}ight)$ Rearranging the terms: $a \left[ 	an A - 	an \left(\frac{A+B}{2}ight) ight] = b \left[ 	an \left(\frac{A+B}{2}ight) - 	an B ight]$ Using $	an x - 	an y = \frac{\sin(x-y)}{\cos x \cos y}$: $a \frac{\sin(A - \frac{A+B}{2})}{\cos A \cos \frac{A+B}{2}} = b \frac{\sin(\frac{A+B}{2} - B)}{\cos B \cos \frac{A+B}{2}}$ $a \frac{\sin(\frac{A-B}{2})}{\cos A} = b \frac{\sin(\frac{A-B}{2})}{\cos B}$ $\sin \left(\frac{A-B}{2}ight) \left( \frac{a}{\cos A} - \frac{b}{\cos B} ight) = 0$ Since $\frac{a}{\sin A} = \frac{b}{\sin B} = k$,we have $a = k \sin A$ and $b = k \sin B$: $\sin \left(\frac{A-B}{2}ight) \left( \frac{k \sin A}{\cos A} - \frac{k \sin B}{\cos B} ight) = 0$ $\sin \left(\frac{A-B}{2}ight) (	an A - 	an B) = 0$ This implies $\sin \left(\frac{A-B}{2}ight) = 0$ or $	an A = 	an B$. In both cases,$A = B$.

If in $\triangle ABC$,$a \tan A + b \tan B = (a + b) \tan \left(\frac{A+B}{2}\right)$,then which of the following holds?

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