Question #cb109

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Question #cb109

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Answer 1 · 2017-03-01T01:05:28

We apply the identities $cot θ = \frac{cos θ}{sin θ}$ $cottheta = costheta/sintheta$ and $csc θ = \frac{1}{sin θ}$ $csctheta = 1/sintheta$ , to obtain:

$\frac{\frac{cos x}{sin x}}{\frac{1}{sin x} - sin x} = sec x$ $(cosx/sinx)/(1/sinx - sinx) = secx$

$\frac{\frac{cos x}{sin x}}{\frac{1 - {sin}^{2} x}{sin x}} = sec x$ $(cosx/sinx)/((1 - sin^2x)/sinx) = secx$

$\frac{cos x}{sin x} \cdot \frac{sin x}{1 - {sin}^{2} x} = sec x$ $cosx/sinx*sinx/(1 - sin^2x) = secx$

Now we can apply the identity ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta + sin^2theta = 1$ :

$\frac{cos x}{sin x} \cdot \frac{sin x}{{cos}^{2} x} = sec x$ $cosx/sinx * sinx/cos^2x = secx$

Eliminate:

$\frac{1}{cos x} = sec x$ $1/cosx = secx$

And this is true by the reciprocal identity $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ .

Our identity has been proved :).

Hopefully this helps!

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Question #cb109

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