## 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:

sec

^{3}(x) - sec^{2}(x) - sec(x) + 1We tried a couple of different approaches, such as factoring sec(x) from each term:

sec(x) * [ sec

^{2}(x) - sec(x) - 1 + 1/sec(x) ]and factoring sec

^{2}(x) from each term:sec

^{2}(x) * [ sec(x) - 1 - 1/sec(x) + 1/sec^{2}(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:

sec

^{3}(x) - sec^{2}(x) - sec(x) + 1We can rearrange the grouping of the terms, which is allowed by the Commutative Property of Addition:

sec

^{3}(x) - sec(x) - sec^{2}(x) + 1Looking at these two groups of terms, we can see that sec(x) can be factored from the first group.

sec(x) * ( sec

^{2}(x) - 1 ) - sec^{2}(x) + 1Now 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 "-sec

^{2}(x) + 1":sec(x) * ( sec

^{2}(x) - 1 ) - 1 * ( sec^{2}(x) - 1 )Now we have the same expression factored from both of the groups of two terms! Let's pull this factor out:

( sec

^{2}(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:

( sec

^{2}(x) - 1 ) * ( sec(x) - 1 )We can look at the first term (sec

^{2}(x) - 1) and see that it is the difference of squares; it is equivalent to (sec^{2}(x) - 1^{2}). 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:

( sec

^{2}(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, tan

^{2}(θ) + 1 = sec^{2}(θ), can easily be arranged to show that:tan

^{2}(θ) = sec^{2}(θ) - 1We now see that tan

^{2}(x) can be substituted in for the first term in our equation:tan

^{2}(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?

Thanks for reading and good luck with your studies,

-Skyler

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