If $\sin \theta=-\frac{1}{2}$ and $\tan \theta=1 / \sqrt{3}$, then $\theta=$

Question

If $\sin \theta=-\frac{1}{2}$ and $\tan \theta=1 / \sqrt{3}$, then $\theta=$, $2 \mathrm{n} \pi+\pi / 6$, $2 \mathrm{n} \pi+11 \pi / 6$, $2 \mathrm{n} \pi+7 \pi / 6$, $2 \mathrm{n} \pi+\pi / 4$

MARKS App · Accepted Answer

We shall first consider values of $\theta$ between 0 $\begin{aligned} & \text { and } 2 \pi \sin \theta=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right) \ & \text { or } \sin (2 \pi-\pi / 6) \ & \therefore \theta=7 \pi / 6 ; 11 \pi / 6 \ & \tan \theta=1 / \sqrt{3}=\tan (\pi / 6) \text { or } \tan (\pi+\pi / 6) \ & \therefore \theta=\pi / 6,7 \pi / 6 \end{aligned}$ The value of $\theta$ which satisfies both the equations is $7 \pi / 6$ Hence the general value of $\theta$ is $2 n \pi+7 \pi / 6$ where $\mathrm{n} \in \mathrm{I}$

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