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

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Shwetank Mauria Share
Mar 16, 2018

Answer:

#y=tantheta# has a period of #pii# and vertical asymptotes are at #theta=((2n+1)pi)/2#

Explanation:

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

Further as #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#

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