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

Answer 1

(b) $\tan 2\alpha =\tan \beta$

$$\begin{gathered} \mathrm{tan\alpha }=\frac{1-\mathrm{cos\beta }}{\mathrm{sin\beta }} \\ =\frac{2{\sin }^2{\displaystyle \frac{\mathrm{\beta }}{2}}}{2\sin {\displaystyle \frac{\mathrm{\beta }}{2}}\cos {\displaystyle \frac{\mathrm{\beta }}{2}}} \\ =\frac{\sin {\displaystyle \frac{\mathrm{\beta }}{2}}}{\cos {\displaystyle \frac{\mathrm{\beta }}{2}}} \\ \Rightarrow \mathrm{tan\alpha }=\tan \frac{\mathrm{\beta }}{2} \\ \Rightarrow \mathrm{\alpha }=\frac{\mathrm{\beta }}{2} \\ \Rightarrow 2\mathrm{\alpha }=\mathrm{\beta } \\ \therefore \tan 2\mathrm{\alpha }=\mathrm{tan\beta } \end{gathered}$$