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Question
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
Options
\[\tan 3 \alpha = \tan 2 \beta\]
- \[\tan 2 \alpha = \tan \beta\]
- \[\tan 2 \alpha = \tan \alpha\]
none of these
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Solution
\[ = \frac{2 \sin^2 \frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}\]
\[ = \frac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}\]
\[ \Rightarrow tan\alpha = \tan\frac{\beta}{2}\]
\[ \Rightarrow \alpha = \frac{\beta}{2}\]
\[ \Rightarrow 2\alpha = \beta\]
\[ \therefore \tan2\alpha = tan\beta\]
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