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Video: Simplifying Trigonometric Expressions Using Pythagorean Identities

Alex Cutbill

Simplify (1 βˆ’ tan 𝜃)² + (1 + tan 𝜃)².

03:38

Video Transcript

Simplify one minus tan πœƒ, squared, plus one plus tan πœƒ, squared.

The first thing we do is expand the brackets; one minus tan πœƒ squared becomes one minus two tan πœƒ plus tan squared πœƒ; and one plus tan πœƒ all squared becomes one plus two tan πœƒ plus tan squared πœƒ.

Now we can collect some like terms. We have two terms of one, so that becomes two. We have a minus two tan πœƒ and a plus two tan πœƒ, which cancel out; they form a zero pair. And finally we have to two tan squared πœƒβ€™s.

Okay, so we have two plus two tan squared πœƒ now, which is certainly simpler than the expression we started with. But I think we can go further. It’s possible that you will notice this is two times one plus tan squared πœƒ. And you would write it in this way because you remember that one plus tan squared πœƒ is equal to sec squared πœƒ.

And so you could write this as two sec squared πœƒ. But what if you hadn’t thought to write two plus two tan squared πœƒ in that way? Well there’s a more reliable method, which is to write everything in terms of sin πœƒ and cos πœƒ. So using the fact that tan πœƒ is sin πœƒ over cos πœƒ, we can rewrite tan squared πœƒ as sin πœƒ over cos πœƒ, squared.

And so the whole expression becomes two plus two times sin squared πœƒ over cos squared πœƒ, which we can now write over a common denominator. So we’ve used the common denominator of cos squared πœƒ, writing it is one fraction.

So now it might be slightly easier to see which identity to apply. We’ve got two cos squared πœƒ plus two sine squared πœƒ, and we know the cos squared πœƒ plus sin squared πœƒ is one, which means as our numerator two cos squared πœƒ plus two sin squared πœƒ is just two.

If you didn’t notice that immediately you might have noticed that sin squared πœƒ can be written in terms of cos squared πœƒ using this identity slightly rearranged. So you can subtract cos squared πœƒ from both sides and get sin squared πœƒ equals one minus cos squared πœƒ. And then substitute that in; you get the same numerator, two.

The final thing is to write to over cos squared πœƒ as two times sec squared πœƒ, using the fact that sec πœƒ is just one over cos πœƒ. So that’s our final answer.

Of course we could have skipped a few lines of working by using the identity one plus tan squared πœƒ is equal to sec squared πœƒ.

The advantage of writing everything in terms of cos πœƒ and sin πœƒ is that although you may take a few more steps to find the final answer, you only have to remember one identity to simplify, which is cos squared πœƒ plus sin squared πœƒ equals one rather than about six identities including one plus tan squared πœƒ equal sec squared πœƒ.