# Interesting problem from a section on Trigonometric Identities

I was working with a student today, and as we worked through the section in his book dealing with Trigonometric Identities and Pythagorean Identities, we stumbled across a problem that gave us a bit of trouble. The solution is not so complicated, but it sure had us stumped earlier.

The problem was presented as such:

Factor and simplify the following using Trigonometric and Pythagorean Identities:

sec3(x) - sec2(x) - sec(x) + 1

We tried a couple of different approaches, such as factoring sec(x) from each term:

sec(x) * [ sec2(x) - sec(x) - 1 + 1/sec(x) ]

and factoring sec2(x) from each term:

sec2(x) * [ sec(x) - 1 - 1/sec(x) + 1/sec2(x) ]

We followed these approaches through a few steps, but nothing we were attempting led to the solution. After doing some reading online, I found that the solution required a simple approach that I had overlooked.

Instead of attempting to draw a factor from all four terms at once, we draw one factor from two of the terms and another factor from the other two terms. So from our original problem:

sec3(x) - sec2(x) - sec(x) + 1

We can rearrange the grouping of the terms, which is allowed by the Commutative Property of Addition:

sec3(x) - sec(x)      - sec2(x) + 1

Looking at these two groups of terms, we can see that sec(x) can be factored from the first group.

sec(x) * ( sec2(x) - 1 )      - sec2(x) + 1

Now it becomes apparent that we can relate the second group of terms to the factoring we've just performed with a little trick: factoring -1 from the expression "-sec2(x) + 1":

sec(x) * ( sec2(x) - 1 )     - 1 * ( sec2(x) - 1 )

Now we have the same expression factored from both of the groups of two terms! Let's pull this factor out:

( sec2(x) - 1 ) * ( sec(x) - 1 )

We now have two approaches we can take to further simplify this problem.

Method 1, Factoring the Difference of Squares:

( sec2(x) - 1 ) * ( sec(x) - 1 )

We can look at the first term (sec2(x) - 1) and see that it is the difference of squares; it is equivalent to (sec2(x) - 12). Knowing this, we could factor the expression one step further to be left with:

(sec(x) + 1)(sec(x) - 1)(sec(x) - 1)
(sec(x) + 1)(sec(x) - 1)2

Method 2, Relating to a Pythagorean Identity:

( sec2(x) - 1 ) * ( sec(x) - 1 )

We could alternatively take a look at this equation and recognize the presence of a Pythagorean Identity.  The second Pythagorean Identity, tan2(θ) + 1 = sec2(θ), can easily be arranged to show that:

tan2(θ) = sec2(θ) - 1

We now see that tan2(x) can be substituted in for the first term in our equation:

tan2(x) * ( sec(x) - 1 )

While either of these methods is correct, it is my personal opinion that factoring the difference of squares is the more elegant expression. What do you think? Is there another approach we could have used? A more elegant solution?

-Skyler

I think that using the difference of squares is more elegant and is the preferable method. This is a convincing example of showing that there are different ways of getting to an answer in math, contrary to the belief of many math teachers (sigh): "It's my way or the highway." Thanks for sharing.
Adrian makes a very good point. As a mathematics teacher, I ALWAYS accept a student's answer provided all of the steps are shown and there are no mistakes. Very few problems have only one way of providing a correct solution. I encourage my students to solve any problem using the method that makes the most sense to them. I will illustrate a way to solve a problem followed by challenging the students to find other ways. You might be surprised by some of the techniques used by the students to find the solution to the problem. I place the students work both inside and outside the classroom using large white paper with an adhesive strip on the top. In this way, my students proudly display their work to whoever walks past or visits the classroom.
I just wanted to add that the beginning factoring type is usually called "Factoring by Grouping."  One thing that I'm a little unsure of though, is why the need to put the original terms in a different order?  You could factor "sec2(x)"  out of the original first two terms, and a "-1" out of the last two terms, as I see it. \$30p/h

Skyler H.

Math whiz here! On call to guide you to success!

20+ hours