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Select the equivalent of
(1sec2α−cos2α+1csc2α−sin2α)⋅cos2αsin2α.\left(\frac{1}{\sec^{2}\alpha-\cos^{2}\alpha}+\frac{1}{\csc^{2}\alpha-\sin^{2}\alpha}\right)\cdot\cos^{2}\alpha\sin^{2}\alpha.(sec2α−cos2α1+csc2α−sin2α1)⋅cos2αsin2α.
This problem is part of the set Trigonometry.
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