Hey, everyone. In this problem, we're asked to verify the identity by working with one side of the equation. Now looking at this equation that we have, we have 1−sinθ over cosθ − cosθ1+sinθ = 0. So it's very clear what side we should be working with, this left side, because we can't do anything to that 0 on the right. So, let's go ahead and get started and take a look at our strategies here. Now looking at these two fractions that are being subtracted, remember that we want to go ahead and add any fractions using a common denominator. Now, I know these fractions are being subtracted, but subtraction is basically just like adding a negative, so we still want to go ahead and combine these fractions with that common denominator. Now looking at the denominators that I have here, I have the cosθ and 1+sinθ. So my common denominator will take both of these and multiply them together to give me a new denominator of cosθ×(1+sinθ). Now, in order to get that common denominator, I need to go ahead and multiply each of these fractions in order to get that common denominator. Now for that common denominator, I need to multiply this first fraction by 1+sinθ. And if I'm multiplying the bottom by that, I also need to multiply the top by that, 1+sinθ. Now multiplying that out will end up giving me a numerator of 1−sin2θ because this is a difference of squares. Now, let's look at that second fraction. We want to multiply both of these by the cosθ in order to get that common denominator. Now on the top, I am left with that minus cos2θ. Remember that we are subtracting here, so that's why I have that minus. Now where do we go from here? Well, remember, we want to be constantly scanning for identities, and I see a couple of different things happening here. Now in my numerator, I have this 1−sin2θ. And coming over here to my identities, I know that my first Pythagorean identity tells me that sin2θ+cos2θ=1. So if I subtract the sin2θ from both sides, it will cancel on that left side and leave me with that 1−sin2θ, which is exactly what I have highlighted here in my numerator. So that one minus sin squared is equal to the cosine squared of theta. So now in my numerator using that Pythagorean identity, I am left with the cos2θ−cos2θ. And then all of that is over that same denominator, the cosθ×(1+sinθ). Now what's happening here? Well, in my numerator, I have the cos2θ−cos2θ. So what happens when I take something and I subtract that same something? Well, I'm just left with 0. Now that right side of my equation is also 0, so I have successfully verified this identity because 0 is equal to 0 and we're done here. Let me know if you have any questions and thanks for watching.