What is the value of $$\frac{[\tan (90 - A) + \cot (90 - A)]^2}{[2 \sec^2 (90 - 2A)]}$$?

$$\frac{[\tan(90-A)+\cot(90-A)]^2}{[2\sec^2(90-2A)]}$$

$$=\frac{[\cot(A)+\tan(A)]^2}{\frac{2}{\sin^2\left(2A\right)}}\ .$$

$$=\frac{\left[\frac{\cos(A)}{\sin\left(A\right)}+\frac{\sin(A)}{\cos\left(A\right)}\right]^2}{\frac{2}{4\left(\sin A\ \cos A\right)^2}}\ .$$

$$=2\left[\frac{\cos(A)}{\sin\left(A\right)}+\frac{\sin(A)}{\cos\left(A\right)}\right]^2\left(\sin\left(A\right)\cos\left(A\right)\right)^2\ .$$

$$=2\left[\cos^2(A)+\sin^2(A)\right]^2\ .$$

$$=2\left(1\right)^2\ .$$

$$=2\ .$$

C is correct choice.