How do you prove $\frac{1 + sec [θ]}{tan [θ] + sin [θ]} = csc [θ]$ $(1+sec[theta])/(tan[theta]+sin[theta])= csc[theta]$ ?

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Answer 1 · 2016-05-03T20:02:43

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Left Side: $= \frac{1 + sec θ}{tan θ + sin θ}$ $=(1+sec theta)/(tan theta +sin theta)$

$= \frac{1 + \frac{1}{cos θ}}{\frac{sin θ}{cos θ} + sin θ}$ $=(1+1/cos theta)/(sin theta /cos theta + sin theta)$

$= \frac{\frac{cos θ + 1}{cos θ}}{\frac{sin θ + sin θ cos θ}{cos θ}}$ $=((cos theta+1)/cos theta)/((sin theta + sin theta cos theta)/cos theta)$

$= \frac{cos θ + 1}{cos θ} \times \frac{cos θ}{sin θ + sin θ cos θ}$ $=(cos theta+1)/cos theta xx cos theta/(sin theta + sin theta cos theta)$

$= \frac{cos θ + 1}{sin θ + sin θ cos θ}$ $=(cos theta+1)/(sin theta + sin theta cos theta)$

$= \frac{cos θ + 1}{sin θ (1 + cos θ)}$ $=(cos theta+1)/(sin theta(1+cos theta))$

$= \frac{cos θ + 1}{sin θ (cos θ + 1)}$ $=(cos theta+1)/(sin theta(cos theta+1))$

$= \frac{1}{sin θ}$ $=1/sin theta$

$= csc θ$ $=csc theta$

$=$ $=$ Right Side

How do you prove 1+sec[θ]tan[θ]+sin[θ]=csc[θ](1+sec[theta])/(tan[theta]+sin[theta])= csc[theta]?