What is \sqrt{(\frac{1+\sin \theta}{\cos \theta})^{2}+(\frac{\cos \theta}{1+\sin \theta})^{2}-2} equal to, where 0 \lt \theta \lt \frac{\pi}{2}?

Explanation

The expression inside the root is of the form a^2 + b^2 - 2ab where a = \frac{1+\sin\theta}{\cos\theta} = \sec\theta+\tan\theta and b = \frac{\cos\theta}{1+\sin\theta} = \sec\theta-\tan\theta. This simplifies to \sqrt{(a-b)^2} = |a-b|. Thus, |\sec\theta+\tan\theta - (\sec\theta-\tan\theta)| = |2\tan\theta| = 2\tan\theta.

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