Question #5d572

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Question #5d572

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Answer 1 · 2016-12-26T07:23:29

Using the definitions of $csc$ $csc$ and $cot$ $cot$ , along with the identities

$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x) = 2sin(x)cos(x)$
$cos (2 x) = 2 {cos}^{2} (x) - 1$ $cos(2x) = 2cos^2(x)-1$

we have

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

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

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

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

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

$= cot (x)$ $=cot(x)$

Answer 2 · 2016-12-26T07:40:12

See proof below

Explanation:

We use

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$

$cos 2 x = 2 {cos}^{2} x - 1$ $cos2x=2cos^2x-1$

$cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$

So,

$csc 2 x + cot 2 x = \frac{1}{sin 2 x} + \frac{cos 2 x}{sin 2 x}$ $csc2x+cot2x=1/(sin2x)+(cos2x)/(sin2x)$

$= \frac{1 + cos 2 x}{sin 2 x}$ $=(1+cos2x)/(sin2x)$

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

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

$= \frac{cos x}{sin x}$ $=cosx/sinx$

$= cot x$ $=cotx$

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Question #5d572

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