\[\begin{aligned} \frac{1}{1-\frac{1}{1-\frac{1}{1-\sec^2x}}} &=\frac{1}{1-\frac{1}{1-\frac{1}{-\tan^2x}}}\\ &=\frac{1}{1-\frac{1}{1+\cot^2x}}\\ &=\frac{1}{1-\frac{1}{\cosec^2x}}\\ &=\frac{1}{1-\sin^2x}\\ &=\frac{1}{\cos^2x}\\ \frac{1}{1-\frac{1}{1-\frac{1}{1-\sec^2x}}}&=\sec^2x \end{aligned}\]
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