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Trigonometry! #2

Geometry Level 3

If $\sec \theta - \tan \theta = \frac {1}{p}$ , then $\csc \theta$ is equal to:

This problem is part of the set Trigonometry.

\frac {p^{2} + 1}{p^{2} - 1}

\frac {2p^{2} - 1}{2p^{2} + 1}

\frac {p^{2} - 1}{p^{2} + 1}

\frac {2p^{2} + 1}{2p^{2} - 1}

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by Omkar Kulkarni

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