What is (1+\cot \alpha-\text{cosec } \alpha)(1+\tan \alpha+\sec \alpha) equal to?

Explanation

Rewriting in sines and cosines gives \frac{\sin \alpha + \cos \alpha - 1}{\sin \alpha} \times \frac{\cos \alpha + \sin \alpha + 1}{\cos \alpha}. This simplifies to \frac{(\sin \alpha + \cos \alpha)^2 - 1}{\sin \alpha \cos \alpha} = \frac{2 \sin \alpha \cos \alpha}{\sin \alpha \cos \alpha} = 2.

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