#### Answer

$\pm\sqrt {1+cot^{2}\theta}$

#### Work Step by Step

By Identity, $csc^{2}\theta=1+cot^{2}\theta$.
Taking the square-root on both sides gives:
$$csc\theta=\pm\sqrt{1+cot^{2}\theta}$$

Published by
Prentice Hall

ISBN 10:
0133186024

ISBN 13:
978-0-13318-602-4

$\pm\sqrt {1+cot^{2}\theta}$

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