Question

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

Answer 1

Apply a Pythagorean Identity and a couple factoring techniques to simplify the expression to `sin^2x`.

Explanation:

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

Let's start with the numerator:
`sec^4x-1`

Note that this can be rewritten as:
`(sec^2x)^2-(1)^2`

This fits the form of a difference of squares, `a^2-b^2=(a-b)(a+b)`, with `a=sec^2x` and `b=1`. It factors into:
`(sec^2x-1)(sec^2x+1)`

From the identity `1+tan^2x=sec^2x`, we can see that subtracting `1` from both sides gives us `tan^2x=sec^2x-1`. We can therefore replace `sec^2x-1` with `tan^2x`:
`(sec^2x-1)(sec^2x+1)`
`->(tan^2x)(sec^2x+1)`

Let's check out the denominator:
`sec^4x+sec^2x`

We can factor out a `sec^2x`:
`sec^4x+sec^2x`
`->sec^2x(sec^2x+1)`

There isn't much more we can do here, so let's look at what we have now:
`((tan^2x)(sec^2x+1))/((sec^2x)(sec^2x+1))`

We can do some canceling:
`((tan^2x)cancel((sec^2x+1)))/((sec^2x)cancel((sec^2x+1))`
`->tan^2x/sec^2x`

Now we rewrite this using only sines and cosines and simplify:
`tan^2x/sec^2x`
`->(sin^2x/cos^2x)/(1/cos^2x)`
`->sin^2x/cos^2x*cos^2x`
`->sin^2x/cancel(cos^2x)*cancel(cos^2x)=sin^2x`