How do I solve this question? If cos θ = -5/13 and 0 ≤ θ ≤ 2π. Determine tan θ I've been stuck on this question and I dont know how to go about this.
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Use the identities $\tan^2{\theta} = \sec^2{\theta}-1$ and $\sec{\theta} = \dfrac{1}{\cos{\theta}}$
Method 2 ... sketch reference triangles in quadrants II and III $\cos{\theta} = \dfrac{adjacent}{hypotenuse}$ $\tan{\theta} = \dfrac{opposite}{adjacent}$
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