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Question
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
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Solution
Given:
\[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
\[ \Rightarrow \text{ tan } A = \frac{2 \sin^2 \frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}} \left( 1 - \cos2\theta = 2 \sin^2 \theta \text{ and } \sin2\theta = 2\sin\theta\cos\theta \right)\]
\[ \Rightarrow \text{ tan } A = \frac{\sin\frac{B}{2}}{\cos\frac{B}{2}} = \tan\frac{B}{2}\]
\[ \Rightarrow A = \frac{B}{2}\]
\[\Rightarrow 2A = B\]
\[ \therefore \tan2A = \text{ tan } B\]
Hence, the value of tan2A is tanB.
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