Question

What are the asymptotes of `g(x)=sec 2x`?

Answer 1

By rewriting a bit,

`g(x)=sec2x=1/{cos2x}`.

There will be vertical asymptotes when the denominator becomes 0, and `cos2x` becomes zero when

`2x=pi/2+npi={2n+1}/2pi` for all integer `n`,

so, by dividing by 2,

`Rightarrow x={2n+1}/4pi`

Hence, the vertical asymptotes are

`x={2n+1}/4pi`

for all integer `n`.

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I hope that this was helpful.