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]
So we need to begin with the definition of tan, we know that [tex] tan x = \frac{sin x}{cos x}[/tex] and we have 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) = sin a sin b + cos a cos b[/tex]
And if we apply those identities into 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)
We can divide numerator and denominator from expression (2) by [tex] \frac{1}{cos x cos y}[/tex] like 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 if we simplity we got:
[tex] tan(x-y) = \frac{tan x - tan y}{tan x tan y +1 }[/tex]
And with that we complete the proof. And that appies for all [tex] (x-y) \neq \frac{\pi}{2} +n\pi[/tex]
