# Find tanθ, if secθ = square root of ten divided by three and sinθ < 0.

**Solution:**

If sec θ = square root of ten divided by three and sin θ < 0

sec θ = (√10)/3 and sin θ < 0

We know that,

cos θ = 1 / sec θ

cos θ = 1/[(√10)/3]

cos θ = (3/√10)

From the trigonometric identity,

sin^{2}θ = 1 - cos^{2}θ

By substituting the value of cos θ,

sin^{2}θ = 1 - (3/√10)^{2}

sin^{2}θ = [1 - (9/10)]

By further calculation,

sin^{2}θ = [(10 - 9) / 10]

sin^{2}θ = 1/10

So we get,

sin θ = +(1/√10), -(1/√10)

sin θ < 0 [Given]

Take the value of sin θ = - (1/√10)

tan θ = (sin θ / cos θ)

tan θ = [ - (1/√10)] / (3/√10)

tanθ = -1/3

Therefore, the value of tan θ is -1/3.

## Find tanθ, if secθ = square root of ten divided by three and sinθ < 0.

**Summary:**

If sec θ = square root of ten divided by three and sin θ < 0, then the value of tan θ is -1/3.

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