Question Video: Simplifying Trigonometric Expressions Using Trigonometric Identities Mathematics • First Year of Secondary School

Video Transcript

Simplify tan 𝜃 sin 𝜃 over sec 𝜃.

To solve this problem, we’re actually gonna have to use a couple of trigonometric identities. The first is that sec 𝜃, just short for secant 𝜃, is actually the reciprocal function of cos 𝜃. So therefore, we can say that sec 𝜃 is equal to one over cos 𝜃. And the second identity that we’re gonna use is actually this one that says that tan 𝜃 is equal to sin 𝜃 over cos 𝜃. And this one is a very common one that you will use in a lot of different problems.

First thing we’re gonna do to simplify our expression, is actually substitute the values that we have from our trig identities into our expression. First, we can see that we’ve substituted in sin 𝜃 over cos 𝜃 for tan 𝜃. And now, we’ve substituted in one over cos 𝜃 for sec 𝜃. So we now have this expression that says that sin 𝜃 over cos 𝜃 multiplied by sin 𝜃 all over one over cos 𝜃 is equal to our original expression of tan 𝜃 sin 𝜃 over sec 𝜃.

Okay, well this is a bit clumsy. So let’s see if I can simplify this a bit. So it can help us to simplify it. So I’ve made it easier to actually understand. We have sin 𝜃 over cos 𝜃, then multiplied by sin 𝜃. And then because we were dividing by one over cos 𝜃, we’ve actually found the reciprocal. And actually, it’s cos 𝜃 over one. And instead of dividing, we’re multiplying. So now, this is much easier to deal with. And we could now fully simplify. Well, as we can see here, I’ve tidied it up. Haven’t quite fully simplified yet, but it’s equal to sin squared 𝜃 cos 𝜃 all over cos 𝜃.

Okay, so one more step. So then if we divide our sin squared 𝜃 cos 𝜃 by cos 𝜃, we’re just left with sin squared 𝜃. So we could say that if we simplify tan 𝜃 sin 𝜃 over sec 𝜃 fully, then we’re left with sin squared 𝜃.

I just want to quickly recap what we did. Just gonna remind you of this kind of problem. First of all, we had a look at some of our trigonometric identities. And we picked these two here to help us with this problem. Next, we used these to substitute back into our expression, to actually eliminate some of the trig ratios we actually had within it. So that gave us this expression over here. We then simplified the expression we had. And it gave us our final answer of sin squared 𝜃.