Question #ccc87

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Question #ccc87

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Answer 1 · 2016-09-22T03:45:50

$\frac{cos θ}{1 + sin θ} + \frac{1 + sin θ}{cos θ}$ $costheta/(1+sintheta)+(1+sintheta)/costheta$

$= \frac{cos θ (1 - sin θ)}{(1 - sin θ) (1 + sin θ)} + \frac{1 + sin θ}{cos θ}$ $=(costheta(1-sintheta))/((1-sintheta)(1+sintheta))+(1+sintheta)/costheta$

$= \frac{cos θ (1 - sin θ)}{1 - {sin}^{2} θ} + \frac{1 + sin θ}{cos θ}$ $=(costheta(1-sintheta))/(1-sin^2theta)+(1+sintheta)/costheta$

$= \frac{cos θ (1 - sin θ)}{{cos}^{2} θ} + \frac{1 + sin θ}{cos θ}$ $=(costheta(1-sintheta))/cos^2theta+(1+sintheta)/costheta$

$= \frac{1 - sin θ}{cos θ} + \frac{1 + sin θ}{cos θ}$ $=(1-sintheta)/costheta+(1+sintheta)/costheta$

$= \frac{1}{cos θ} - \frac{sin θ}{cos θ} + \frac{1}{cos θ} + \frac{sin θ}{cos θ}$ $=1/costheta-sintheta/costheta+1/costheta+sintheta/costheta$

$= \frac{2}{cos θ} = 2 sec θ$ $=2/costheta=2sectheta$

Alternative

$\frac{cos θ}{1 + sin θ} + \frac{1 + sin θ}{cos θ}$ $costheta/(1+sintheta)+(1+sintheta)/costheta$

$= \frac{{cos}^{2} θ + {(1 + sin θ)}^{2}}{(1 + sin θ) cos θ}$ $=(cos^2theta+(1+sintheta)^2)/((1+sintheta)costheta)$

$= \frac{{cos}^{2} θ + (1 + 2 sin θ + {sin}^{2} θ)}{(1 + sin θ) cos θ}$ $=(cos^2theta+(1+2sintheta+sin^2theta))/((1+sintheta)costheta)$

$= \frac{2 + 2 sin θ}{(1 + sin θ) cos θ}$ $=(2+2sintheta)/((1+sintheta)costheta)$

$= \frac{2}{cos θ} = 2 sec θ$ $=2/costheta=2sectheta$

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Question #ccc87

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