Find the exact value of sec theta given that sin theta = -2/5 and cos theta > 0.

\(\sin(\theta) < 0 \mbox{ and }\cos(\theta)>0\)

 

\(\sin(\theta) = -\dfrac 2 5 \Rightarrow \cos(\theta) = \pm \sqrt{1-\sin^2(\theta)}\\ \mbox{We know }\cos(\theta)>0 \mbox{ so we select the positive value.}\\ \cos(\theta) = \sqrt{1-\dfrac{4}{25}} = \dfrac{\sqrt{21}}{5}\)

 

\(\sec(\theta)=\dfrac 1 {\cos(\theta)} = \dfrac {5}{\sqrt{21}}= \dfrac{5\sqrt{21}}{21}\)

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