Hey, everyone. In this example problem, we're asked to use our Pythagorean identities in order to rewrite the expression with no fraction. And the expression that we're given here is 11+cosθ. It might not be immediately apparent how we can use our Pythagorean identities here because we don't even have a squared trig function, but let's play around with this a little bit.
In my denominator, I have 1+cosθ. Let's see what happens if I multiply this by 1-cosθ. Remember we're working with a fraction, so if I'm multiplying the denominator, let's see what happens when I multiply this out here. In my numerator, I of course just have 1-cosθ, having multiplied that by 1. But in my denominator, 1+cosθ1-cosθ, using my difference of squares, I get that this is 1-cos2θ.
Now I have a squared trig function and I can recognize this as being just a different form of that first Pythagorean identity. So, if I subtract cos2θ from both sides of this Pythagorean identity, it leaves me with sin2θ=1-cos2θ, which is the exact expression that I have here. This is a sort of trick if you want to get a Pythagorean identity when you don't have one. If you have 1+sometrigfunction, if you multiply it by 1-thatsametrigfunction, you'll end up with a Pythagorean identity.
Using that here, I can replace that denominator with sin2θ. And in my numerator, I still just have 1-cosθ. But remember, our goal was here, we don't want a fraction. So how can we get rid of this fraction? Let's think about this. In this expression, I could rewrite this as being 1-cosθsin2θ. But remember, one over the sine of theta is just the cosecant. So one over the sine squared is simply csc2θ. So now I'm just left with csc2θ1-cosθ. No more fraction. So this is my final answer here that11+cosθ is equal to csc2θ1-cosθ. Let me know if you have any questions. Thanks for watching, and I'll see you in the next one.