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:
(b) $\tan 2 \alpha=\tan \beta$
$\tan \alpha=\frac{1-\cos \beta}{\sin \beta}$
$=\frac{2 \sin ^{2} \frac{\beta}{2}}{2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}}$
$=\frac{\sin \frac{\beta}{2}}{\cos \frac{\beta}{2}}$
$\Rightarrow \tan \alpha=\tan \frac{\beta}{2}$
$\Rightarrow \alpha=\frac{\beta}{2}$
$\Rightarrow 2 \alpha=\beta$
$\therefore \tan 2 \alpha=\tan \beta$
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