If `tanA=(1-cosB)/(sinB)` , then `tan2A=tanBdot`

Explanation

To solve the problem where \( \tan A = \frac{1 - \cos B}{\sin B} \) and we need to prove that \( \tan 2A = \tan B \), we will follow these steps:

### Step 1: Rewrite \( 1 - \cos B \)
We can use the identity \( 1 - \cos B = 2 \sin^2 \frac{B}{2} \). Thus, we rewrite the expression for \( \tan A \):
\[ \tan A = \frac{2 \sin^2 \frac{B}{2}}{\sin B} \]

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