# How do you simplify (sec^4x-1)/(sec^4x+sec^2x)?

Jul 14, 2016

Apply a Pythagorean Identity and a couple factoring techniques to simplify the expression to ${\sin}^{2} x$.

#### Explanation:

Recall the important Pythagorean Identity $1 + {\tan}^{2} x = {\sec}^{2} x$. We will be needing it for this problem.

${\sec}^{4} x - 1$

Note that this can be rewritten as:
${\left({\sec}^{2} x\right)}^{2} - {\left(1\right)}^{2}$

This fits the form of a difference of squares, ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$, with $a = {\sec}^{2} x$ and $b = 1$. It factors into:
$\left({\sec}^{2} x - 1\right) \left({\sec}^{2} x + 1\right)$

From the identity $1 + {\tan}^{2} x = {\sec}^{2} x$, we can see that subtracting $1$ from both sides gives us ${\tan}^{2} x = {\sec}^{2} x - 1$. We can therefore replace ${\sec}^{2} x - 1$ with ${\tan}^{2} x$:
$\left({\sec}^{2} x - 1\right) \left({\sec}^{2} x + 1\right)$
$\to \left({\tan}^{2} x\right) \left({\sec}^{2} x + 1\right)$

Let's check out the denominator:
${\sec}^{4} x + {\sec}^{2} x$

We can factor out a ${\sec}^{2} x$:
${\sec}^{4} x + {\sec}^{2} x$
$\to {\sec}^{2} x \left({\sec}^{2} x + 1\right)$

There isn't much more we can do here, so let's look at what we have now:
$\frac{\left({\tan}^{2} x\right) \left({\sec}^{2} x + 1\right)}{\left({\sec}^{2} x\right) \left({\sec}^{2} x + 1\right)}$

We can do some canceling:
((tan^2x)cancel((sec^2x+1)))/((sec^2x)cancel((sec^2x+1))
$\to {\tan}^{2} \frac{x}{\sec} ^ 2 x$

Now we rewrite this using only sines and cosines and simplify:
${\tan}^{2} \frac{x}{\sec} ^ 2 x$
$\to \frac{{\sin}^{2} \frac{x}{\cos} ^ 2 x}{\frac{1}{\cos} ^ 2 x}$
$\to {\sin}^{2} \frac{x}{\cos} ^ 2 x \cdot {\cos}^{2} x$
$\to {\sin}^{2} \frac{x}{\cancel{{\cos}^{2} x}} \cdot \cancel{{\cos}^{2} x} = {\sin}^{2} x$