Respuesta :
Answer:
See the proof below
Step-by-step explanation:
For this case we need to proof the following identity:
[tex] tan(x-y) = \frac{tan(x) -tan(y)}{1+ tan(x) tan(y)}[/tex]
We need to begin with the definition of tangent:
[tex] tan (x) =\frac{sin(x)}{cos(x)}[/tex]
So we can replace into our formula and we got:
[tex] tan(x-y) = \frac{sin(x-y)}{cos(x-y)}[/tex] (1)
We have the following identities useful for this case:
[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]
If we apply the identities into our equation (1) we got:
[tex] tan(x-y) = \frac{sin(x) cos(y) - sin(y) cos(x)}{sin(x) sin(y) + cos(x) cos(y)}[/tex] (2)
Now we can divide the numerator and denominato from expression (2) by [tex] \frac{1}{cos(x) cos(y)}[/tex] and 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{sin(x) sin(y)}{cos(x) cos(y)} +\frac{cos(x) cos(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 this identity is satisfied for all:
[tex] (x-y) \neq \frac{\pi}{2} +n\pi [/tex]
