MCQ
If $\tan\alpha=\frac{1-\cos\beta}{\sin\beta},$ then:
- A$\tan3\alpha=\tan2\beta$
- B$\tan2\alpha=\tan\beta$
- C$\tan2\beta=\tan\alpha$
- DNone of these
Solution:
$\tan\alpha=\frac{1-\cos\beta}{\sin\beta}$
$=\frac{2\sin^2\frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}$
$=\frac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}$
$\Rightarrow\tan\alpha=\tan\frac{\beta}{2}$
$\Rightarrow\alpha=\frac{\beta}{2}$
$\Rightarrow2\alpha=\beta$
$\therefore\tan2\alpha=\tan\beta$
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