# How do you verify tan^2(x) / sec(x) = sec(x) - cos(x) ?

Aug 22, 2015

First prove that $\sec x - \cos x = \left(\sin x\right) \left(\tan x\right)$. Then, mess with the equation some to get the right answer.

#### Explanation:

Start with the identity ${\sin}^{2} x + {\cos}^{2} x = 1$

Subtract ${\cos}^{2} x$ from both sides:

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

Divide both sides by $\cos x$.

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

Now simplify. Remember sin/cos = tan, and 1/cos = sec

$\sin x \cdot \sin \frac{x}{\cos} x = \sec x - \cos x$

$\sin x \cdot \tan x = \sec x - \cos x$

There. Now that we've proven that $\left(\sin x\right) \left(\tan x\right) = \sec x - \cos x$, all we have to do is show that ${\tan}^{2} \frac{x}{\sec} x = \left(\sin x\right) \left(\tan x\right)$.

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

There we go! If this doesn't make sense, don't worry. Let me know in the 'comments' and I'll give you a list of the trigonometric identities I used to solve this.