Respuesta :
Given:
f(θ) = tan(θ) = 3
Note that
[tex]tan(x+y) = \frac{tanx + tany}{1-tanx \, tany} [/tex]
Note that tan(π) = 0.
Therefore
f(θ + π) = tan(θ+π)
= (tanθ + tan π)/(1 -tanθ tan π)
= 3/1 = 3
Answer: 3
f(θ) = tan(θ) = 3
Note that
[tex]tan(x+y) = \frac{tanx + tany}{1-tanx \, tany} [/tex]
Note that tan(π) = 0.
Therefore
f(θ + π) = tan(θ+π)
= (tanθ + tan π)/(1 -tanθ tan π)
= 3/1 = 3
Answer: 3
As given
[tex]f(\theta) = tan( \theta) = 3[/tex]
So [tex]f(\theta + \pi) = tan (\theta + \pi)[/tex]
And we know [tex]tan(\theta + \pi) = -tan(\theta)[/tex] because it is in second quadrant and tan is negative in second quadrant.
So [tex]f(\theta + \pi) = -tan(\theta) = -3[/tex]
So answer is -3.
[tex]f(\theta) = tan( \theta) = 3[/tex]
So [tex]f(\theta + \pi) = tan (\theta + \pi)[/tex]
And we know [tex]tan(\theta + \pi) = -tan(\theta)[/tex] because it is in second quadrant and tan is negative in second quadrant.
So [tex]f(\theta + \pi) = -tan(\theta) = -3[/tex]
So answer is -3.
