In a right ­angled triangle ABC, ∠B is the right angle and AC = 2√5­ cm. If AB - BC = 2 cm then the value of $$(cos^2 A - cos^2 C)$$ is :

By Pythagoras theorem,
$$x^2 + (x-2)^2 = 20$$
$$x^2 + x^2 +4x+4 =20$$
$$2x^2+4x+4 =20$$
$$x^2+2x+2=10$$
Solving the quadratic equation we get
x=4 and x=-2
Seince x cannot be negative x=4.
AC= $$2 \sqrt{5}$$
$$Cos A = \frac{x}{2 \sqrt{5}}$$
$$Cos ^{2} A = \frac{x^{2}}{20}= \frac{16}{20}=\frac{4}{5}$$

$$Cos C =Cos (90-A) =Sin A = \frac{x-2}{2 \sqrt{5}}$$

$$Sin ^{2} A = \frac{(x-2)^{2}}{20}=\frac{4}{20}=\frac{1}{5}$$

$$(cos^2 A - cos^2 C) = Cos ^{2} A - Sin ^{2} A = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}$$
Hence Option B is the correct answer.