If 64^{\sin^2 \theta}+64^{\cos^2 \theta}=16 where 0 \leq \theta \leq \frac{\pi}{2} then what is the value of \tan \theta+\cot \theta?

Explanation

Let x = 64^{\sin^2 \theta}. Since \cos^2 \theta = 1 - \sin^2 \theta, we have x + \frac{64}{x} = 16. Solving x^2 - 16x + 64 = 0 yields x = 8. Thus 64^{\sin^2 \theta} = 8 \implies 2^{6 \sin^2 \theta} = 2^3 \implies \sin^2 \theta = \frac{1}{2}, making \theta = 45^\circ. The value of \tan 45^\circ + \cot 45^\circ = 1 + 1 = 2.

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