If \( \tan\alpha=\frac{1-\cos\beta}{\sin\beta} \), Then Which Relation is True?
Question
If
\[ \tan\alpha=\frac{1-\cos\beta}{\sin\beta}, \]
then
(a) \(\tan 3\alpha=\tan 2\beta\)
(b) \(\tan 2\alpha=\tan\beta\)
(c) \(\tan 2\beta=\tan\alpha\)
(d) none of these
Solution
Use the standard half-angle identity:
\[ \tan\frac{\beta}{2} = \frac{1-\cos\beta}{\sin\beta} \]
Given,
\[ \tan\alpha = \frac{1-\cos\beta}{\sin\beta} \]
Therefore,
\[ \tan\alpha = \tan\frac{\beta}{2} \]
Hence,
\[ \alpha=\frac{\beta}{2} \]
Multiplying both sides by 2,
\[ 2\alpha=\beta \]
Taking tangent on both sides,
\[ \tan 2\alpha = \tan\beta \]
Final Answer
\[ \boxed{\tan 2\alpha=\tan\beta} \]
Therefore, the correct option is (b).