Question #d1e3c

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Question #d1e3c

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Answer 1 · 2016-12-04T21:29:20

Put on a common denominator and use the identity $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$ .

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

Use the identity ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$ .

$\frac{1 + 2 sin x + {sin}^{2} x + {cos}^{2} x}{cos x + cos x sin x} = \frac{2}{cos x}$ $(1 + 2sinx + sin^2x + cos^2x)/(cosx + cosxsinx) = 2/cosx$

$\frac{1 + 2 sin x + 1}{cos x + cos x sin x} = \frac{2}{cos x}$ $(1 + 2sinx + 1)/(cosx + cosxsinx) = 2/cosx$

$\frac{2 + 2 sin x}{cos x + cos x sin x} = \frac{2}{cos x}$ $(2 + 2sinx)/(cosx + cosxsinx) = 2/cosx$

$\frac{2 (1 + sin x)}{cos x (1 + sin x)} = \frac{2}{cos x}$ $(2(1 + sinx))/(cosx(1 + sinx)) = 2/cosx$

$\frac{2}{cos x} = \frac{2}{cos x}$ $2/cosx = 2/cosx$

Hopefully this helps!

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