Question #d7a8e

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Question #d7a8e

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Answer 1 · 2017-05-25T01:28:24

See Below.

Explanation:

Identity Required:
${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x) +cos^2(x)=1$
${sin}^{2} (x) = 1 - {cos}^{2} (x)$ $sin^2(x)=1-cos^2(x)$

Now solving:
$\frac{sin (t)}{1 + cos (t)} = \frac{1 - cos (t)}{sin (t)}$ $sin(t)/(1+cos(t))=(1-cos(t))/sin(t)$
$\frac{sin (t)}{1 + cos (t)} \cdot \frac{1 - cos (t)}{1 - cos (t)} = R H S$ $sin(t)/(1+cos(t))*(1-cos(t))/(1-cos(t))=RHS$
$\frac{sin (t) - sin (t) cos (t)}{1 - {cos}^{2} (t)} = R H S$ $(sin(t)-sin(t)cos(t))/(1-cos^2(t))=RHS$
$\frac{sin (t) - sin (t) cos (t)}{{sin}^{2} (t)} = R H S$ $(sin(t)-sin(t)cos(t))/sin^2(t)=RHS$
$\frac{1 - cos (t)}{sin (t)} = R H S$ $(1-cos(t))/sin(t)=RHS$
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Question #d7a8e

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