If p=\sec \theta-\tan \theta and q=\text{cosec } \theta+\cot \theta, then what is p+q(p-1) equal to?

Explanation

Let \theta = 45^\circ. Then p = \sqrt{2} - 1 and q = \sqrt{2} + 1. Substitute into p + q(p-1) = (\sqrt{2}-1) + (\sqrt{2}+1)(\sqrt{2}-2) = \sqrt{2}-1 + 2 - 2\sqrt{2} + \sqrt{2} - 2 = -1.

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