**Proof:**

Let's start by expanding the expression (a - bcos2A)(a - bcos2B):

(a - bcos2A)(a - bcos2B) = a^2 - abcos2B - abcos2A + b^2cos2Acos2B

To simplify this further, we need to express cos2A and cos2B in terms of tanA and tanB.

Recall the trigonometric identity: 1 + tan^2x = sec^2x

From this identity, we can rewrite:

cos^2x = 1 - sin^2x
= 1 - (1/tan^2x)
= (tan^2x - 1) / tan^2x

Using this identity, we can express cos2A and cos2B:

cos2A = (tan^2A - 1) / tan^2A
cos2B = (tan^2B - 1) / tan^2B

Now, let's substitute these values into the expression we obtained earlier:

(a - bcos2A)(a - bcos2B) = a^2 - ab(tan^2B - 1) / tan^2B - ab(tan^2A - 1) / tan^2A + b^2[(tan^2A - 1) / tan^2A][(tan^2B - 1) / tan^2B]

To simplify this further, we can multiply through by tan^2A and tan^2B to remove the denominators:

(a - bcos2A)(a - bcos2B) = a^2tan^2A - ab(tan^2B - 1) - ab(tan^2A - 1) + b^2(tan^2A - 1)(tan^2B - 1)

Expanding further, we get:

(a - bcos2A)(a - bcos2B) = a^2tan^2A - abtan^2B + ab - abtan^2A + ab + b^2tan^2A - b^2 - b^2tan^2B + b^2

Combining like terms, we obtain:

(a - bcos2A)(a - bcos2B) = a^2tan^2A - abtan^2B - abtan^2A + b^2tan^2A - b^2tan^2B + a^2

Rearranging the terms, we have:

(a - bcos2A)(a - bcos2B) = a^2(tan^2A + tan^2B) - b^2(tan^2A + tan^2B) + (a^2 - b^2)

Since tan^2A + tan^2B = (a - b)/(ab) (given in the question), we can substitute this value:

(a - bcos2A)(a - bcos2B) = a^2[(a - b)/(ab)] - b^2[(a - b)/(ab)] + (a^2 - b^2)

Simplifying further, we get:

(a - bcos2A)(a - bcos2B) = (