If \text{cosec } \theta - \sin \theta = m and \sec \theta - \cos \theta = n, then what is m^{\frac{4}{3}}n^{\frac{2}{3}} + m^{\frac{2}{3}}n^{\frac{4}{3}} equal to?

Explanation

We can write m = \frac{1-\sin^2\theta}{\sin\theta} = \frac{\cos^2\theta}{\sin\theta} and n = \frac{1-\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}. The given expression is m^{\frac{2}{3}}n^{\frac{2}{3}}(m^{\frac{2}{3}} + n^{\frac{2}{3}}). Here m^{\frac{2}{3}}n^{\frac{2}{3}} = \cos^{\frac{2}{3}}\theta \sin^{\frac{2}{3}}\theta. Substituting these into the bracket simplifies perfectly to \sin^2\theta + \cos^2\theta = 1.

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