# How do you simplify \frac { 1} { \sin x + 1} + \frac { 1} { \sin x - 1}?

Mar 26, 2017

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

#### Explanation:

First get a common denominator by multiplying the first term by $1$ in the form of $\frac{\sin x - 1}{\sin x - 1}$ and the second in the form of $\frac{\sin x + 1}{\sin x + 1}$

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

Multiply out

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

And combine

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

Combine like terms

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

And you're done!

Mar 27, 2017
• 2 tan x.sec x

#### Explanation:

Put the 2 fractions having common denominator
$f \left(x\right) = \frac{2 \sin x}{{\sin}^{2} x - 1} = \frac{2 \sin x}{-} \left(1 - {\sin}^{2} x\right) = - \frac{2 \sin x}{\cos} ^ 2 x =$
$f \left(x\right) = - 2 \tan x \left(\frac{1}{\cos} x\right) = - 2 \tan x . \sec x$