Given csc ∅ = 13/5, find sec 0 · tan 0.

Do use mean sec and tan of theta (θ)? This is what I think you mean.

If cscθ = 13/5, sinθ = 5/13 because cscθ = 1/sinθ.

We also know that...

cos²θ + sin²θ = 1

So...

cos²θ + (5/13)² = 1

cos²θ = 1-(5/13)²

cos²θ = 1 - 25/169

cos²θ = 144/169

cosθ = 12/13

tanθ = sinθ/cosθ

tanθ = (5/13)/(12/13) = 5/12

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step-by-step explanation:

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