Hey everyone. In this problem, we're asked to verify the identity. And the equation that we're given here is the cosine of a plus b over the cosine of a times the cosine of b, and that's equal to the negative tangent of a times the tangent of b plus 1. Now remember that there are always going to be multiple ways to go about verifying an equation being equal, that's totally fine. Feel free to try this on your own and then check back in with me. Here, I'm gonna show you how I would do this. Now remember that when verifying an identity, we want to start with our more complicated side. And here, this fraction appears more complicated to me than this right side, so I'm going to start on that left side with that fraction. Now what do we want to do here? Well, remember one of our strategies tells us that we want to be constantly scanning for identities. And the first thing that I notice here is that in my numerator, I have the cosine of a plus b.
cos a + b cos × cos − sin × sin cos × cosNow since we're adding in that argument, we're gonna be subtracting those two terms from each other. So let's go ahead and expand this out using that sum formula. So I'll end up with the cosine of a times the cosine of b minus the sine of a times the sine of b. Now that denominator stays the same here, the cosine of a times the cosine of b. Now we always want to be constantly checking in with what's actually going on in our expression. And looking at this expression, one thing that I notice is that in my numerator, I have this term, the cosine of a times the cosine of b. And that's the exact same term that's in my denominator. So I'm gonna do something here that seems a little bit backwards because typically we want to go ahead and combine all of our fractions. But here, since these terms are the same, I'm actually going to break this fraction up. So let's see what happens when I do that. Now breaking this fraction up, I have the cosine of a times the cosine of b over the cosine of a times the cosine of b minus the sine of a times the sine of b over the cosine of a times the cosine of b. So what happened here? Well, looking at this term, these are the exact same term in the numerator and the denominator. So this is just equal to 1. Then looking at this term over here, let's break this up further and see what happens.
1 − sin cos × sin cosSo here, I'm subtracting the sine of a over the cosine of a, and this is being multiplied by the sine of b over the cosine of b. Now what is sine of something over the cosine of that same something? Well, it's the tangent. So here, I really have a 1 minus the tangent of a times the tangent of b. So why did I want to get here? Well, remember what the right side of my equation is, because our goal here is to make both sides of our equation equal. Now that right side of my equation is negative tangent of a times the tangent of b plus 1, and that's exactly what this is if I rearrange it. So let's rearrange this and fully verify this identity. Now rearranging this to look like that right side, this is negative tangent of a times the tangent of b plus 1. Now this is exactly the same as that right side, and I don't have to do anything else to that right side. I don't have to simplify anything. I can just bring it right down here, the tangent of a times the tangent of b plus 1 and we're fully done here. We have successfully verified this identity with both sides of our equation being equal here. Let me know if you have any questions, and thanks for watching.
