Calculus (3rd Edition)

$$y=\pm \sqrt{1+x^2}.$$
We have $x=\tan t$, $y=\sec t$. Use the trig identity: $$1+\tan^2 t=\sec^2t$$ Combine with $x$ and $y$: $$y^2=1+x^2.$$ Solve for $y$: $$y=\pm \sqrt{1+x^2}.$$