Answer: [tex]\sec(\theta) = -\frac{\sqrt{145}}{8}[/tex]
This is the fraction with the square root of 145 over 8. The "8" is not inside the square root. Don't forget about the negative sign out front.
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Work Shown:
[tex]\sec^2(\theta) = 1+\tan^2(\theta)[/tex]
[tex]\sec^2(\theta) = 1+\left(\frac{9}{8}\right)^2[/tex]
[tex]\sec^2(\theta) = 1+\frac{81}{64}[/tex]
[tex]\sec^2(\theta) = \frac{64}{64}+\frac{81}{64}[/tex]
[tex]\sec^2(\theta) = \frac{64+81}{64}[/tex]
[tex]\sec^2(\theta) = \frac{145}{64}[/tex]
[tex]\sec(\theta) = \pm\sqrt{\frac{145}{64}}[/tex]
[tex]\sec(\theta) = -\sqrt{\frac{145}{64}}[/tex] See note below
[tex]\sec(\theta) = -\frac{\sqrt{145}}{\sqrt{64}}[/tex]
[tex]\sec(\theta) = -\frac{\sqrt{145}}{8}[/tex]
note: because theta is in quadrant III, this means that cos(theta) is negative, so sec(theta) is also negative. Recall that sec(theta) = 1/cos(theta) which is one of the reciprocal identies.
