Respuesta :
y=tanx=sinx/cosx we cant divide by 0 so
cosx=0 would be a problm
cosx=0 when x=pi/2,3pi/2,...(2n+1)pi/2
(2n+1)pi/2=(2npi+pi)/2=2npi/2+pi/2=npi+pi/2
so its B
cosx=0 would be a problm
cosx=0 when x=pi/2,3pi/2,...(2n+1)pi/2
(2n+1)pi/2=(2npi+pi)/2=2npi/2+pi/2=npi+pi/2
so its B
The correct answer is: Option (B) [tex]\frac{\pi}{2} + n\pi[/tex]
Explanation:
Given function:
[tex]y = tan(x)[/tex]
We know that,
[tex]tan(x) = \frac{sin(x)}{cos(x)}[/tex]
To find the domain, put the "denominator" equal to zero, as follows:
[tex]cos(x) = 0[/tex]
Now ask yourself the following question to find domain: For what values of x, will [tex]cos(x)[/tex] be equal to 0? Well, for [tex]x = \frac{\pi }{2}, \frac{3\pi}{2},\frac{5\pi}{2},..., \frac{\pi}{2}(1+2n)[/tex] where n = any integer.
Therefore, the domain of y = tan(x) is:
[tex]\frac{\pi}{2}(1+2n)\\\frac{\pi}{2} + \frac{2n\pi}{2}\\[/tex]
[tex]\frac{\pi}{2} + n\pi[/tex] (Option B)