# Question 68cf7

Mar 7, 2017

see below

#### Explanation:

Use the following Properties:

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

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

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

LEFT HAND SIDE:

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

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

=(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)#

$= \frac{\cos 2 \left(\frac{x}{2}\right)}{1}$

$= \frac{\cos \cancel{2} \left(\frac{x}{\cancel{2}}\right)}{1}$

$= \cos x$

$\therefore =$ RIGHT HAND SIDE