If \frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=p \sec \theta+q \tan \theta, where 0 \lt \theta \lt \frac{\pi}{2} then what is p+q equal to?

Explanation

Divide the numerator and denominator by \cos \theta to get \frac{\tan \theta - 1 + \sec \theta}{\tan \theta + 1 - \sec \theta}. Replace the 1 in the numerator with (\sec^2 \theta - \tan^2 \theta). The numerator becomes (\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = (\sec \theta + \tan \theta)(1 - \sec \theta + \tan \theta). Canceling out (1 - \sec \theta + \tan \theta) with the denominator leaves \sec \theta + \tan \theta. Thus, p=1 and q=1, so p+q=2.

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