Question #68cf7

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Question #68cf7

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Answer 1 · 2017-03-07T18:04:00

see below

Explanation:

Use the following Properties:

1. Quotient Property:
$tan x = \frac{sin x}{cos x}$ $tan x = sinx /cos x$

2. Pythagorean Property:
${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2 x =1$

3. Double Argument Property
$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos 2x = cos^2x-sin^2 x$

LEFT HAND SIDE:

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

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

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

$=(cos^2 (x/2)-sin^2(x/2))/cancel(cos^2(x/2))*cancel(cos^2(x/2))/(cos^2(x/2)+sin^2(x/2)$

$=(cos^2 (x/2)-sin^2(x/2))/(cos^2(x/2)+sin^2(x/2)$

$=(cos2(x/2))/1$

$=(coscancel2(x/cancel2))/1$

$= cos x$

$:.=$ RIGHT HAND SIDE

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Question #68cf7

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