Answer:
D. proves [tex]tan(w + \pi) = tanw[/tex]
Step-by-step explanation:
Given
[tex]tan(w + \pi) = tan(w)[/tex]
See attachment for complete question
Option D answers the question (See proof below).
Using tangent sum identity, we have:
[tex]tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB}[/tex]
In this case:
[tex]A = w[/tex]
[tex]B = \pi[/tex]
So, we have:
[tex]tan(w + \pi) = \frac{tanw + tan \pi}{1 - tanw. tan \pi}[/tex]
Note that
[tex]tan\ pi = 0[/tex]
So:
[tex]tan(w + \pi) = \frac{tanw + tan \pi}{1 - tanw. tan \pi}[/tex]
[tex]tan(w + \pi) = \frac{tanw + 0}{1 - tanw * 0}[/tex]
[tex]tan(w + \pi) = \frac{tanw}{1}[/tex]
[tex]tan(w + \pi) = tanw[/tex]
Hence: Option D answers the question
