How do you identify the period and asympotes for $y = tan θ$ $y=tantheta$ ?

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Answer 1 · 2018-03-16T15:27:51

$y = tan θ$ $y=tantheta$ has a period of $π i$ $pii$ and vertical asymptotes are at $θ = \frac{(2 n + 1) π}{2}$ $theta=((2n+1)pi)/2$

As $tan (π + θ) = tan θ$ $tan(pi+theta)=tantheta$ , $y = tan θ$ $y=tantheta$ has a period of $π$ $pi$ .

Further as $θ \to \frac{(2 n + 1) π}{2}$ $theta->((2n+1)pi)/2$ , from left or right i.e.

$lim_(theta->(((2n+1)pi)/2)^+)tantheta=oo$ and

$lim_(theta->(((2n+1)pi)/2)^-)tantheta=-oo$

Hence vertical asymptotes are at $theta=((2n+1)pi)/2$

How do you identify the period and asympotes for y=tanthetay=tanθy=tantheta?