What is the <strong>MINIMUM</strong> value of 3600 \sec^{2}\theta+121 \csc^{2}\theta, where 0 \lt \theta \lt \frac{\pi}{2}?

Explanation

The minimum value of a^2\sec^2\theta + b^2\csc^2\theta is (a+b)^2. Here a^2 = 3600 \implies a=60, and b^2 = 121 \implies b=11. Minimum value = (60+11)^2 = 71^2 = 5041.

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