# Verify sec^2x-tan^2x=tanxcotx by changing only one side?

##### 2 Answers
Feb 23, 2018

${\sec}^{2} x - {\tan}^{2} x = \tan x \cot x$ would validate to 1 = 1.

#### Explanation:

To solve this, we will have to use the Reciprocal Identity and the Pythagorean Identity.

The Reciprocal Identity basically defines $\csc$, $\sec$, $\cot$ from the known $\sin$, $\cos$, and $\tan$, respectively.

The Pythagorean Identity is defined as:

${\cos}^{2} + {\sin}^{2} = 1$

We can manipulate this into two parts:

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

These will come in handy here in a moment.

Knowing these, all we really have to do is change the left side accordingly:

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

This can be further simplified into:

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

$\left({\cos}^{2} / {\cos}^{2}\right) x = \tan x \cot x$

$1 = \tan x \cot x$

From here, we aren't necessarily changing the right side, but we are simplifying it.

Since $\tan x = \left(\frac{\sin}{\cos}\right)$ and $\cot x = \left(\frac{\cos}{\sin}\right)$ they would both cancel out and become one.

Therefore, ${\sec}^{2} x - {\tan}^{2} x = \tan x \cot x$ would validate to 1 = 1.

Feb 23, 2018

Please see below.

#### Explanation:

.

${\sec}^{2} x - {\tan}^{2} x = \tan x \cot x$

${\sec}^{2} x - {\tan}^{2} x = \frac{1}{\cos} ^ 2 x - {\sin}^{2} \frac{x}{\cos} ^ 2 x = \frac{1 - {\sin}^{2} x}{\cos} ^ 2 x = {\cos}^{2} \frac{x}{\cos} ^ 2 x = {\cos}^{2} \frac{x}{\cos} ^ 2 x \cdot \frac{\sin x \cos x}{\sin x \cos x} = \cos \frac{x}{\cos} x \cdot \cos \frac{x}{\cos} x \cdot \sin \frac{x}{\cos} x \cdot \cos \frac{x}{\sin} x = \left(\cos \frac{x}{\cos} x \cdot \sin \frac{x}{\cos} x\right) \left(\cos \frac{x}{\cos} x \cdot \cos \frac{x}{\sin} x\right) = \left(\cancel{\textcolor{red}{\cos}} \frac{x}{\cos} x \cdot \sin \frac{x}{\cancel{\textcolor{red}{\cos}}} x\right) \left(\cos \frac{x}{\cancel{\textcolor{red}{\cos}}} x \cdot \cancel{\textcolor{red}{\cos}} \frac{x}{\sin} x\right) = \left(\sin \frac{x}{\cos} x\right) \left(\cos \frac{x}{\sin} x\right) = \tan x \cot x$