If tan A = frac{1 - cos B}{sin B}, then tan 2 | vedclass

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(A) Given $	an A = \frac{1 - \cos B}{\sin B}$.Using the trigonometric identities $1 - \cos B = 2 \sin^2 \left( \frac{B}{2} ight)$ and $\sin B = 2 \

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$	an B$

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(A) Given $	an A = \frac{1 - \cos B}{\sin B}$.Using the trigonometric identities $1 - \cos B = 2 \sin^2 \left( \frac{B}{2} ight)$ and $\sin B = 2 \sin \left( \frac{B}{2} ight) \cos \left( \frac{B}{2} ight)$,we get:$	an A = \frac{2 \sin^2 \left( \frac{B}{2} ight)}{2 \sin \left( \frac{B}{2} ight) \cos \left( \frac{B}{2} ight)}$$	an A = 	an \left( \frac{B}{2} ight)$.This implies $A = \frac{B}{2}$,which means $2A = B$.Therefore,$	an 2A = 	an B$.

If $\tan A = \frac{1 - \cos B}{\sin B},$ then $\tan 2A =$

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