Answer: [tex]\tan \theta=\dfrac{1}{\sqrt3}.[/tex]
Step-by-step explanation: Given that
[tex]\sin\theta=-\dfrac{1}{2}[/tex] and [tex]\theta[/tex] lies in Quadrant III.
We are to find the value of [tex]\tan \theta.[/tex]
We will be using the following trigonometric identities:
[tex](i)~sin^2\theta+\cos^2\theta=1,\\\\(ii)~\dfrac{\sin\theta}{\cos{\theta}}=\tan \theta.[/tex]
We have
[tex]\tan\theta\\\\\\=\dfrac{\sin\theta}{\cos\theta}\\\\\\=\dfrac{\sin\theta}{\pm\sqrt{1-\sin^2\theta}}\\\\\\=\pm\dfrac{-\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\frac{\sqrt3}{2}}\\\\\\=\pm\dfrac{1}{\sqrt3}.[/tex]
Since [tex]\theta[/tex] lies in Quadrant III, so tangent will be positive.
Thus,
[tex]\tan \theta=\dfrac{1}{\sqrt3}.[/tex]
