# How do you prove  (secx-tanx)(secx+tanx) =secx ?

Dec 13, 2015

The given identity is false.

$\left(\sec \left(x\right) - \tan \left(x\right)\right) \left(\sec \left(x\right) + \tan \left(x\right)\right) = 1$

#### Explanation:

We will be using the following:

• $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ (by definition)

• $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ (by definition)

• $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$ (difference of squares formula)

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

$\left(\sec \left(x\right) - \tan \left(x\right)\right) \left(\sec \left(x\right) + \tan \left(x\right)\right) = {\sec}^{2} \left(x\right) - {\tan}^{2} \left(x\right)$
(by the difference of squares formula)

$= {\left(\frac{1}{\cos} \left(x\right)\right)}^{2} - {\left(\sin \frac{x}{\cos} \left(x\right)\right)}^{2}$
(by definition of secant and tangent)

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

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

$= {\cos}^{2} \frac{x}{\cos} ^ 2 \left(x\right)$
(by the above identity)

$= 1$