If tan B = frac{2 sin A sin C}{sin (A+C)},the | vedclass

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(B) Given $	an B = \frac{2 \sin A \sin C}{\sin (A+C)}$.Taking the reciprocal on both sides,we get $\frac{1}{	an B} = \frac{\sin (A+C)}{2 \sin A \sin

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Harmonic progression

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(B) Given $	an B = \frac{2 \sin A \sin C}{\sin (A+C)}$.Taking the reciprocal on both sides,we get $\frac{1}{	an B} = \frac{\sin (A+C)}{2 \sin A \sin C}$.Using the expansion $\sin (A+C) = \sin A \cos C + \cos A \sin C$,we have $\frac{1}{	an B} = \frac{\sin A \cos C + \cos A \sin C}{2 \sin A \sin C}$.This simplifies to $\frac{1}{	an B} = \frac{1}{2} (\cot A + \cot C) = \frac{1}{2} (\frac{1}{	an A} + \frac{1}{	an C})$.Multiplying by $2$,we get $\frac{2}{	an B} = \frac{1}{	an A} + \frac{1}{	an C}$.This is the condition for $	an A, 	an B, 	an C$ to be in Harmonic Progression.

If $\tan B = \frac{2 \sin A \sin C}{\sin (A+C)}$,then $\tan A, \tan B$ and $\tan C$ are in

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