Question

If sinθ = -1/2 and θ is in Quadrant III, then tanθ = _____.

Answer 1

Answer:  $\tan \theta=\dfrac{1}{\sqrt3}.$

Step-by-step explanation:  Given that

$\sin\theta=-\dfrac{1}{2}$ and $\theta$ lies in Quadrant III.

We are to find the value of $\tan \theta.$

We will be using the following trigonometric identities:

$(i)~sin^2\theta+\cos^2\theta=1,\\\\(ii)~\dfrac{\sin\theta}{\cos{\theta}}=\tan \theta.$

We have

$\tan\theta\\\\\\=\dfrac{\sin\theta}{\cos\theta}\\\\\\=\dfrac{\sin\theta}{\pm\sqrt{1-\sin^2\theta}}\\\\\\=\pm\dfrac{-\frac{1}{2}}{\sqrt{1-\left(\frac{1}{2}\right)^2}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\sqrt{1-\frac{1}{4}}}\\\\\\=\pm\dfrac{\frac{1}{2}}{\frac{\sqrt3}{2}}\\\\\\=\pm\dfrac{1}{\sqrt3}.$

Since $\theta$ lies in Quadrant III, so tangent will be positive.

Thus,

$\tan \theta=\dfrac{1}{\sqrt3}.$

Answer 2

Tan tetha at quadrant III would be $\frac{ \sqrt{3} }{3}$