Recall that \(x=r\cos\theta\). When \(\cos\theta
eq0\), we can write
\(r=\sin\theta\tan\theta\iff r\cos\theta=x=\sin^2\theta\)
You have \(x=1\) whenever \(\theta\) is a multiple of \(\dfrac\pi2\). But \(\cos\dfrac\pi2=0\), which would make \(r\) undefined. So \(x=1\) must be a vertical asymptote.
So assume \(\theta
eq\dfrac{n\pi2}\) for any integer \(n\). Then
\(x=\sin^2\theta\)
and since \(|\sin\theta|<1\) for all \(\theta\), it follows that \(0<\sin^2\theta<1\), or \(0.
