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 \( \frac{1}{1 + \cos(\theta)} \). Now 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 \( 1 + \cos(\theta) \). Now let's see what happens if I multiply this by \( 1 - \cos(\theta) \). Now 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 \( 1 - \cos(\theta) \), having multiplied that by 1. But in my denominator, \( 1 + \cos(\theta) \times 1 - \cos(\theta) \), using my difference of squares, I get that this is \( 1 - \cos^2(\theta) \). 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 \( \cos^2(\theta) \) from both sides of this Pythagorean identity it leaves me with \( \sin^2(\theta) = 1 - \cos^2(\theta) \), which is the exact expression that I have here. So this is a sort of trick if you want to get a Pythagorean identity when you don't have one. If you have \( 1 + \) some trig function, if you multiply it by \( 1 - \) that same trig function, you'll end up with a Pythagorean identity. Now using that here, I can replace that denominator with \( \sin^2(\theta) \). And in my numerator, I still just have that \( 1 - \cos(\theta) \). But remember what our goal was here, we don't want a fraction. So how can we get rid of this fraction? Well let's think about this. In this expression, I could rewrite this as being \( \frac{1}{\sin^2(\theta)} \times (1 - \cos(\theta)) \). And now the only reason that we have a fraction here is this \( \frac{1}{\sin^2(\theta)} \). But remember, \( \frac{1}{\sin(\theta)} \) is just the cosecant. So \( \frac{1}{\sin^2(\theta)} \) is simply the cosecant squared of theta. So now I'm just left with \( \csc^2(\theta) \times (1 - \cos(\theta)) \). No more fraction. So this is my final answer here that \( \frac{1}{1 + \cos(\theta)} = \csc^2(\theta) \times (1 - \cos(\theta)) \). Let me know if you have any questions. Thanks for watching, and I'll see you in the next one.