Respuesta :
Answer:
See the proof below.
Step-by-step explanation:
For this case we need to proof the following indentity:
[tex] tan(x+y) = \frac{tan (x) + tan(y)}{1- tan(x) tan(y)}[/tex]
So we need to begin with the definition of tangent, we know that [tex] tan (x) =\frac{sin(x)}{cos(x)}[/tex] and we can do this:
[tex] tan (x+y) = \frac{sin (x+y)}{cos(x+y)}[/tex] (1)
We also have the following identities:
[tex] sin (a+b) = sin (a) cos(b) + sin (b) cos(a)[/tex]
[tex] cos(a+b)= cos(a) cos(b) - sin(a) sin(b)[/tex]
Now we can apply those identities into equation (1) like this:
[tex]tan (x+y) =\frac{sin (x) cos(y) + sin (y) cos(x)}{cos(x) cos(y) - sin(x) sin(y)}[/tex] (2)
We can divide numerator and denominator from expression (2) by [tex] \frac{1}{cos(x) cos(y)}[/tex] we got this:
[tex] tan (x+y) = \frac{\frac{sin (x) cos(y)}{cos (x) cos(y)} + \frac{sin(y) cos(x)}{cos(x) cos(y)}}{\frac{cos(x) cos(y)}{cos(x) cos(y)} -\frac{sin(x)sin(y)}{cos(x) cos(y)}}[/tex]
And simplifying we got:
[tex] tan (x+y) = \frac{tan(x) + tan(y)}{1-tan(x) tan(y)}[/tex]
And that complete the proof.
