Question

I need help with this equation

Answer 1

Answer:
`\sec(\theta) = -(√(145))/(8)`

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:

`\sec^2(\theta) = 1+\tan^2(\theta)`

`\sec^2(\theta) = 1+\left((9)/(8)\right)^2`

`\sec^2(\theta) = 1+(81)/(64)`

`\sec^2(\theta) = (64)/(64)+(81)/(64)`

`\sec^2(\theta) = (64+81)/(64)`

`\sec^2(\theta) = (145)/(64)`

`\sec(\theta) = \pm\sqrt{(145)/(64)}`

`\sec(\theta) = -\sqrt{(145)/(64)}` See note below

`\sec(\theta) = -(√(145))/(√(64))`

`\sec(\theta) = -(√(145))/(8)`

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.