# Question #4ae15

Apr 24, 2016

Use the facts that:
$\textcolor{w h i t e}{\text{XXX}} \frac{\sin \left(x\right)}{\cos \left(x\right)} = \tan \left(x\right)$
and
$\textcolor{w h i t e}{\text{XXX}} 1 - {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right)$

#### Explanation:

Note that I have assumed that the intended requirement was to show that:
$\textcolor{w h i t e}{\text{XXX}} \frac{\cos \left(x\right)}{1 - \sin \left(x\right)} - \frac{\cos \left(x\right)}{1 + \sin \left(x\right)} = 2 \tan \left(x\right)$
(as expressed in the question the expression is ambiguous)

$\frac{\cos \left(x\right)}{1 - \sin \left(x\right)} - \frac{\cos \left(x\right)}{1 + \sin \left(x\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\left(\cos \left(x\right)\right) \cdot \left(1 + \sin \left(x\right)\right)}{\left(1 - \sin \left(x\right)\right) \cdot \left(1 + \sin \left(x\right)\right)} - \frac{\left(\cos \left(x\right)\right) \cdot \left(1 - \sin \left(x\right)\right)}{\left(1 + \sin \left(x\right)\right) \cdot \left(1 - \sin \left(x\right)\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\left(\cos \left(x\right) + \cos \left(x\right) \cdot \sin \left(x\right)\right) - \left(\cos \left(x\right) - \cos \left(x\right) \cdot \sin \left(x\right)\right)}{1 - {\sin}^{2} \left(x\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{2 \cdot \cos \left(x\right) \cdot \sin \left(x\right)}{{\cos}^{2} \left(x\right)}$

$\textcolor{w h i t e}{\text{XXX}} = 2 \cdot \sin \frac{x}{\cos} \left(x\right)$

$\textcolor{w h i t e}{\text{XXX}} = 2 \tan \left(x\right)$