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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.

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

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Excel in math and science

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by Omkar Kulkarni

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