Hey, everyone. We just learned our sum and difference identities for sine and cosine, so we of course 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 plus b, this is going to be equal to:
tan ( a ) + tan ( b ) 1 - tan ( a ) ⋅ tan ( b )Then for the tangent of a minus b, this is going to be almost identical except now we have:
tan ( a ) - tan ( b ) 1 + tan ( a ) ⋅ tan ( b )Now here you'll notice that whenever we're taking the tangent of a plus b, we're going to be adding in the numerator and subtracting in the denominator. Then when we're taking the tangent of a minus b, we are instead subtracting in the numerator and adding in the denominator. Now let's go ahead and take a look at this example here and apply these new identities. So here we have the tangent of pi + pi over 4. So we want to go ahead and use our sum formula here since these angles are being added. So expanding this out, this gives me:
tan ( π ) + tan ( π 4 ) 1 - tan ( π ) ⋅ tan ( π 4 )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 over 4. Now all I'm left with in that numerator is the tangent of pi over 4 and in my denominator just 1. But the tangent of pi over 4 over 1 is just the tangent of pi over 4. And I know from my unit circle again that the tangent of pi over 4 is simply equal to 1, giving me my final answer: The tangent of pi + pi over 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 over 12 radians in order to find the exact value for trig functions that are not on the unit circle. Now remember when working with these identities, we are also going to come across expressions that have variables in them. So let's go ahead and take a look at this example here. Here we have the tangent of theta minus 45 degrees. 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:
tan ( θ ) - tan ( 45 ° ) 1 + tan ( θ ) ⋅ tan ( 45 ° )The tangent of 45 degrees from our unit circle is just equal to 1. Here I can replace both of those with ones. Now all I have in my numerator is the tangent of theta minus 1. Then in my denominator, I have 1 plus the tangent of theta times 1. 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 minus 1 over 1 plus the tangent of theta. Now here you'll notice that we ended up with just a function of theta, whereas we started with a function of theta minus 45 degrees. Now, 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 degrees...
