Question

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

Answer 1

tan tetha at quadrant III would be
`( √(3) )/(3)`

Answer 2

Answer:
`\tan \theta=(1)/(\sqrt3).`

Step-by-step explanation: Given that

`\sin\theta=-(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)~(\sin\theta)/(cos(\theta))=\tan \theta.`

We have

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

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

Thus,

`\tan \theta=(1)/(\sqrt3).`