If tanθ - cotθ = α and cosθ sinθ = b, then the value of (a^2 + 4) (b^2 - ­1)^2 is:

Sin(2a) = 2sin(a/2)cos(a/2)
Cos(2a) = 2cos$$^2$$a - 1 = 1 - 2sin$$^2$$a
Sin$$^2$$a + cos$$^2$$a = 1
cosec$$^2$$a - cot$$^2$$a = 1
Given,
tanθ - cotθ = a
(sinθ/cosθ)−(cosθ/sinθ)=a
(sin$$^2$$θ−cos$$^2$$θ)/cosθsinθ=a
=−2cos2θ/sin2θ=a
a = -2cot2θ
Also given, cosθ - sinθ = b
Squaring both sides and using (a - b)$$^2$$ = a$$^2$$ + b$$^2$$ - 2ab, we get,
Cos$$^2$$θ + sin$$^2$$θ - 2cosθsinθ = b$$^2$$
1 - sin2θ = b$$^2$$
We have to find the value of
(a$$^2$$ + 4) (b$$^2$$ - 1)$$^2$$
(4cot$$^2$$2θ + 4)(1 - sin2θ - 1)2
4(cosec$$^2$$2θ)(-sin2θ)$$^2$$
= 4
Option A is the correct answer.