# If cosθ = negative two over five and tanθ > 0, what is the value of sinθ?

**Solution:**

cosθ = -2/5, tanθ > 0 (Given)

We should find the value of sin θ.

In Quadrant II and III, cosine is negative.

In Quadrant I and III, tangent is positive.

Thus θ lies in Quadrant III

We shall use the trigonometric identity

cos^{2}θ + sin^{2}θ = 1

As cosθ = -2/5

√(1 - sin^{2}θ) = -2/5

Squaring on both sides

1 - sin^{2}θ = 4/25

sin^{2}θ = 1 - 4/25

sin^{2}θ = 21/5

sinθ = ± √21/5

As the angle lies in Quadrant III and sine is negative in Quadrant III

sinθ = -√21/5

## If cosθ = negative two over five and tanθ > 0, what is the value of sinθ?

**Summary:**

If cosθ = negative two over five and tan θ > 0, the value of sinθ is -√21/5.