# How do you prove 2tan(x)sec(x) = (1/(1-sin(x))) - (1/(1+sin(x)))?

Dec 14, 2015

See explanation...

#### Explanation:

Recall these identities involving trig functions:

Quotient Identity: $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
Reciprocal Identity: $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$
Pythagorean Identity: $1 - {\sin}^{2} \left(x\right) = {\cos}^{2} \left(x\right)$

Using these, we can rewrite the identity:

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

We then make common denominators:

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

And simplify using the Pythagorean Identity from above:

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