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## Precalculus

### Unit 2: Lesson 9

Angle addition identities- Trig angle addition identities
- Using the cosine angle addition identity
- Using the cosine double-angle identity
- Using the trig angle addition identities
- Proof of the sine angle addition identity
- Proof of the cosine angle addition identity
- Proof of the tangent angle sum and difference identities

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# Proof of the tangent angle sum and difference identities

CCSS.Math:

Using the sine and cosine of the sum or difference of two angles, we can prove: tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y)). Created by Sal Khan.

## Want to join the conversation?

- At3:35, why did Sal combine cos(y) and sin(y) to become tan(y)? Hadn't he already canceled the cos(y) out?(2 votes)
- Without showing the step, Sal broke the fraction down into two parts. (sinxcosy+cosxsiny)/(cosxcosy) was separated into (sinxcosy)/(cosxcosy) + (cosxsiny/(cosxcosy). In the first, cosy cancels to get sinx/cosx=tanx and in the second cancel cosx to get siny/cosy=tany. So you canceled cosy in one part and cosx in the second part which allows you to do what your question is about. Hope this helps(4 votes)

- Are there any videos about addition and difference identities of secants,cosecants and cotangents?(2 votes)
- Since you know what the identities for cosine, sine and tangent are, you can figure them out for the inverse functions!

For example, cot(90 + 45) = 1/tan(90 + 45) and so on.(3 votes)

## Video transcript

- [Instructor] In this
video I'm going to assume that you already know a few things and we've covered this. We've proved this in other videos that sine of x plus y is
equal to sine of x cosine y plus and then you swap
the cosines and the sines, cosine of x sine y, and then cosine of x plus y
is equal to cosine x cosine y minus sine x sin y. Once again, we've proven
this in this in other videos and then there's some other
properties we know of cosine and sine that we have
looked at another video's. Cosine of -x is equal to cosine of x and that sine of negative
x is equal to -sine of x. And that of course, the tangent of something is defined as a sine over cosine of that something. Now with that out of the way, I wanna come up with a formula for tangent of x plus y expressed just in terms of tangent of x and tangent of y. You can view it as the antilog for what we did up here
for sine and cosine. Well, the immediate thing
that you might recognize is that tangent of x plus
y based on the definition of tangent is the same
thing as sine of x plus y over cosine of x plus y. And what's that going to be equal to? Well, we know that sine of x plus y can be expressed this way. So let me write that down. So that's going to be sine of x cosine y plus cosine of x sine of y. And then, and actually, so that we can save a
little bit of writing, I'm gonna awkwardly write, make the line down here, because we're gonna put
something here in a second, but I think you'll get the idea. So there's going to be that
over cosine of x plus y which is this expression when
you just express it in terms of cosines of x and
cosines of y and sines of x and sines of y. So let me write it here. So you're gonna have cosine
of x cosine y minus sine of x, and then sine y. Now we wanna express
everything in terms of tangents of xs and ys. And so it might make sense here to say, all right, well, we know
tangent is sine over cosine. So what if we were to
divide both the numerator and the denominator by some
expression that can start to make the numerator
and denominator express in terms of tangents. And I will cut a little
bit to the chase here. So in the numerator, what I can do is, and I'm gonna do this
just in the numerator, and then I'm gonna do it
in the denominator as well. I'm gonna divide the numerator
by cosine of x cosine y. And of course, I can't
just divide the numerator by cosine of x cosine of y that would change the value of the, this rational expression. I have to do that to
the denominator as well. So I know this is a very
complex looking fraction here but it's going to simplify in a second. So I'm also going to
divide the denominator by cosine of x cosine of y. And now let's see if we can simplify this in certain ways. In the numerator, we can see
that this cosine y cancels with this cosine y. And so that first term becomes slightly in another color here. So this sine of x over cosine of x. And so the numerator, I can say this is going
to be equal to sine of x over cosine of x is tangent of x. And then the second term here, we can see that this cosine of x cancels with this cosine of x . So we're left with sine
of y over cosine of y, which is of course, tangent of y. So plus tangent of y and then all of that is going to be over, now we can look at the denominator. So this first term here, we can see the cosine of x
cancels with the cosine of x and the cosine of y cancels
out with the cosine of y. So you could view this first
term here when you divide by this cosine of x cosine y, it just becomes one and then we're going to have the minus. And now this second term is interesting. We have sine of x over cosine of x, sine of y over cosine of y. So sine of x over cosine of x that over there is tangent of x, and then sine of y over
cosine of y its tangent of y. So this is going to be tangent
of x times tangent of y. And just like that, we have come up with an expression for tangent of x plus y that just deals with tangent of xs and tangent of ys. Now the next question you might say, well, all right, that's
great for tangent of x plus y but what about tangent of x minus y? Well, here we just have
to recognize a little bit of what we've seen before. Let me write it over here. Tangent of -x is equal to sine of -x over cosine of negative x and what's that going to be equal to? And I know I'm running out of space. This is going to be equal to
sine of -x is the same thing as -sine of x, -sine of x, and then cosine of -x is just cosine of x. Well, this is just the
negative of the tangent of x. So this is negative tangent of x. And the reason why that is useful is I can rewrite this as being, write here. This is the same thing
as tangent of x plus -y. So everywhere we saw a y here, we can replace it with a -y. So this is going to be
equal to tangent of x plus the tangent of -y, all of that over 1 minus the tangent of x times the tangent of -y. Well, we know the tangent of -y is the same thing as the
negative tangent of y. And we know that over here as well. So this could just be, we could write the tangent of y here, and then the negative would
turn this into a plus. And so just to write everything neatly, we know that also the tangent of x minus y can be rewritten as tangent
of x minus tangent of y all of that over 1 plus
tangent of x and tangent of y.