Hey, everyone. We just learned our sum and difference identities for sine and cosine, so of course, we can't forget about tangent. Now these identities are not going to look quite as nice, but we're still going to use them for the same purpose, to simplify expressions and to find the exact value of functions. So let's not waste any time here and jump right into our sum and difference identities for tangent.
Now for the tangent of \(a + b\) this is going to be equal to the tangent of \(a\) plus the tangent of \(b\) over \(1 - \) the tangent of \(a \times\) the tangent of \(b\). Then for the tangent of \(a - b\), this is going to be almost identical except now we have the tangent of \(a\) minus the tangent of \(b\) over \(1 +\) the tangent of \(a \times\) the tangent of \(b\). Here you'll notice that whenever we're taking the tangent of \(a + b\), we're adding in the numerator and subtracting in the denominator. Then when we're taking the tangent of \(a - b\), we are instead subtracting in the numerator and adding in the denominator.
Let's go ahead and take a look at this example here and apply these new identities. Here we have the tangent of \(\pi + \pi/4\). So we want to go ahead and use our sum formula here since these angles are being added. Expanding this out, this gives me the tangent of that first angle \(\pi\) plus the tangent of that second angle \(\pi/4\). And this gets divided by one minus the tangent of that first angle, again, \(\pi\), times the tangent of \(\pi/4\). Now we know these angles from the unit circle. Right? And we know that the tangent of \(\pi\) is simply equal to 0. So that term goes away in the numerator, and then in my denominator, this entire term goes away as well because it's the tangent of \(\pi\) multiplying the tangent of \(\pi/4\). Now all I'm left with in the numerator is the tangent of \(\pi/4\) and in my denominator just 1. But the tangent of \(\pi/4\) over 1 is just the tangent of \(\pi/4\). And I know from my unit circle again that the tangent of \(\pi/4\) is simply equal to 1, giving me my final answer. The tangent of \(\pi + \pi/4\) is equal to 1, having used my sum identity there.
Now remember from our sine and cosine sum and difference identities that we want to use these whenever our argument contains a plus or a minus. And we also want to use these whenever our argument contains an angle that's a multiple of 15 degrees or \(\pi/12\) radians in order to find the exact value for trig functions that are not on the unit circle. When working with these identities, we are also going to come across expressions that have variables in them. Let's take a look at this example here. Here we have the tangent of \(\theta - 45^\circ\). So let's go ahead and expand this out using our difference formula since these angles are being subtracted. Here expanding this out I end up with the tangent of that first angle \(\theta\) minus the tangent of that second angle 45 degrees, and this is getting divided by 1 plus the tangent of \(\theta\) times the tangent of 45 degrees. Now we know the tangent of 45 degrees from our unit circle even if we don't know the tangent of \(\theta\) because that's just a variable. But the tangent of 45 degrees is just equal to 1. So here I can replace both of those with ones. Now all I have in my numerator is the tangent of \(\theta - 1\). Then in my denominator, I have \(1 + \) the tangent of \(\theta\). But the tangent of \(\theta \times 1\) is just the tangent of \(\theta\), so this gives me my final simplified expression here, the tangent of \(\theta - 1\) over \(1 + \) the tangent of \(\theta\). Here you'll notice that we ended up with just a function of \(\theta\), whereas we started with a function of \(\theta - 45^\circ\). This is something that you'll be asked to do from time to time, and you can do this using your sum and difference formulas.
Let's take a look at one final example here. Here we have the tangent of \(90^\circ + \theta\). Let's go ahead and use our sum formula to expand this out. Here I end up with the tangent of \(90^\circ\) plus the tangent of \(\theta\) over \(1 -\) the tangent of \(90^\circ \times\) the tangent of \(\theta\). At this point, you might be thinking, okay. I know the tangent of \(90^\circ\) from my unit circle. But thinking of what that value actually is, the tangent of \(90^\circ\) is undefined. So here, we would end up with undefined plus the tangent of \(\theta\) over \(1 -\) another undefined value. So how do we deal with that here? Well, whenever you end up with the tangent of \(a\) or the tangent of \(b\) being undefined, you're going to start over and you're going to rewrite the tangent of \(a +\) or \( - b\) as the sine over the cosine. So here with the tangent of \(90^\circ + \theta\), I would rewrite this as the sine of \(90^\circ + \theta\) over the cosine of \(90^\circ + \theta\), and then go from there using my sum and difference formulas for sine and cosine rather than for tangent, because that way I won't end up with an undefined value.
Now that we know how to use our sum and difference identities for tangent, let's continue practicing. Thanks for watching and let me know if you have questions.