If \frac{\cos \theta}{\csc \theta+1}+\frac{\cos \theta}{\csc \theta-1}=2 where 0 \lt \theta \lt 90^\circ, then what is the value of \sin^4\theta+\cos^4\theta?

Explanation

Simplifying the left hand side: \cos\theta \left[ \frac{\csc\theta-1 + \csc\theta+1}{\csc^2\theta-1} \right] = \frac{\cos\theta (2\csc\theta)}{\cot^2\theta} = \cos\theta \cdot \frac{2}{\sin\theta} \cdot \frac{\sin^2\theta}{\cos^2\theta} = 2\tan\theta. So, 2\tan\theta = 2 \implies \tan\theta = 1 \implies \theta = 45^\circ. The value is \sin^4(45^\circ) + \cos^4(45^\circ) = (\frac{1}{\sqrt{2}})^4 + (\frac{1}{\sqrt{2}})^4 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.

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