Solution
$$(\frac{secA}{cotA + tanA})^{2}$$
=$$(\frac{secA}{\frac{1}{tanA} + tanA})^{2}$$
=$$(\frac{secA}{\frac{sec^{2}A}{tanA}})^{2}$$
=$$(sinA)^{2}$$
=$$1-cos^{2}A$$
so the answer is option A.
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