How do you factor sec^2(x)-sec(x)+sin^2(x)*sec^2(x)?

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How do you factor sec^2(x)-sec(x)+sin^2(x)*sec^2(x)?

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Answer 1 · 2018-06-14T19:07:43

${sec}^{2} x - sec x + {sin}^{2} x {sec}^{2} x = - {sec}^{2} x (cos x + 2) (cos x - 1)$ $sec^2x-secx+sin^2xsec^2x=-sec^2x(cosx+2)(cosx-1)$

Explanation:

Start by factoring out $- {sec}^{2} x$ $-sec^2x$ to get
$- {sec}^{2} x (- 1 + \frac{1}{sec x} - {sin}^{2} x)$ $-sec^2x(-1+1/secx-sin^2x)$
Using the fact that $sec x = \frac{1}{cos x}$ $secx=1/cosx$ and ${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$ , we get
$- {sec}^{2} x (- 1 + cos x - 1 + {cos}^{2} x)$ $-sec^2x(-1+cosx-1+cos^2x)$
Rearranging gives
$- {sec}^{2} x ({cos}^{2} x - cos x - 2)$ $-sec^2x(cos^2x-cosx-2)$
Recognizing that this is a quadratic expression of $cos x$ $cosx$ , we can factor to get
$- {sec}^{2} x (cos x + 2) (cos x - 1)$ $-sec^2x(cosx+2)(cosx-1)$

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How do you factor sec^2(x)-sec(x)+sin^2(x)*sec^2(x)?

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