If \sin\theta=\frac{12}{13}, then what is the value of (\tan\theta+\sec\theta)^{2}(\csc\theta-\cot\theta)^{-2}, 0 \lt \theta \lt \frac{\pi}{2}?

Explanation

From \sin\theta = \frac{12}{13}, we get a 5-12-13 triangle where \cos\theta = \frac{5}{13}. Thus, \tan\theta = \frac{12}{5}, \sec\theta = \frac{13}{5}, \csc\theta = \frac{13}{12}, \cot\theta = \frac{5}{12}. The first term is (\frac{12}{5} + \frac{13}{5})^2 = 5^2 = 25. The second term is (\frac{13}{12} - \frac{5}{12})^{-2} = (\frac{8}{12})^{-2} = (\frac{2}{3})^{-2} = \frac{9}{4}. The product is 25 \times \frac{9}{4} = \frac{225}{4}.

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