# How do you simplify Sinx/(1-secx) - sin/(1+secx)?

Apr 19, 2016

We must remember that $\sec = \frac{1}{\cos}$

#### Explanation:

$\sin \frac{x}{1 - \frac{1}{\cos} x} - \sin \frac{x}{1 + \frac{1}{\cos} x}$

$\sin \frac{x}{\frac{\cos x - 1}{\cos} x} - \sin \frac{x}{\frac{\cos x + 1}{\cos} x}$

$\sin x \times \cos \frac{x}{\cos x - 1} - \sin x \times \cos \frac{x}{\cos x + 1}$

$\frac{\sin x \cos x}{\cos x - 1} - \frac{\sin x \cos x}{\cos x + 1}$

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

$\frac{\sin x {\cos}^{2} x + \sin x \cos x}{{\cos}^{2} x - 1} - \frac{\sin x {\cos}^{2} x - \sin x \cos x}{{\cos}^{2} x - 1}$

$\frac{\sin x {\cos}^{2} x + \sin x \cos x - \sin x {\cos}^{2} x + \sin x \cos x}{{\cos}^{2} x - 1}$

$\frac{2 \sin x \cos x}{{\cos}^{2} x - 1}$

Applying the double angle identity $2 \sin x \cos x = \sin 2 x$ and the pythagorean identity ${\cos}^{2} x - 1 = - {\sin}^{2} x$ we get the following:

$\frac{\sin 2 x}{- {\sin}^{2} x} \to$ this is your answer.

*Beware: This is not the only possible answer; there are many ways of attacking this problem, with different answers completely within the realm of possibility. *

Hopefully this helps!