#### Answer

$$ y=\pm \sqrt{1+x^2}.$$

#### Work Step by Step

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 $y$:
$$ y=\pm \sqrt{1+x^2}.$$

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Published by
W. H. Freeman

ISBN 10:
1464125260

ISBN 13:
978-1-46412-526-3

$$ y=\pm \sqrt{1+x^2}.$$

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