Question #781b9

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Question #781b9

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Answer 1 · 2016-11-29T03:40:32

We will use the following:

$cot (x) = \frac{cos (x)}{sin (x)}$ $cot(x) = cos(x)/sin(x)$
$csc (x) = \frac{1}{sin (x)}$ $csc(x) = 1/sin(x)$
${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x) = 1$
$\Rightarrow {cos}^{2} (x) = 1 - {sin}^{2} (x)$ $=> cos^2(x) = 1-sin^2(x)$
$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2 = (a+b)(a-b)$

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

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

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

$= \frac{1 - {sin}^{2} (x)}{sin (x) (1 + sin (x))}$ $=(1-sin^2(x))/(sin(x)(1+sin(x)))$

$= \frac{(1 + sin (x)) (1 - sin (x))}{sin (x) (1 + sin (x))}$ $=((1+sin(x))(1-sin(x)))/(sin(x)(1+sin(x)))$

$= \frac{1 - sin (x)}{sin (x)}$ $=(1-sin(x))/sin(x)$

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Question #781b9

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