Proof:
To prove the given identity, we need to simplify the left-hand side and then equate it to the right-hand side.

Simplification of L.H.S:
We have tan^3 theta/1 × cot^3 theta/ 1
= (tan theta/ cot theta)^3
= (1/ tan theta × cos theta/ sin theta)^3
= (cos theta/ sin theta)^3 × (sin theta/ cos theta)^3
= 1/ sin^3 theta × cos^3 theta
= 1/ (sin theta × cos theta)^3
= 1/ (sin theta cos theta)^3

Also, we have tan^2 theta/ 1 × cot^2 theta/ 1
= (tan theta/ cot theta)^2
= (1/ tan theta × cos theta/ sin theta)^2
= (cos theta/ sin theta)^2 × (sin theta/ cos theta)^2
= 1/ sin^2 theta × cos^2 theta
= 1/ (sin theta cos theta)^2

Therefore, L.H.S = (1/ (sin theta cos theta)^3) × (1/ (sin theta cos theta)^2)
= 1/ (sin theta cos theta)^5

Simplification of R.H.S:
We have sec theta × cosec theta - 2 sin theta cos theta
= 1/ cos theta × 1/ sin theta - 2 sin theta cos theta
= (1/ (sin theta cos theta)) - 2 sin theta cos theta
= 1/ (sin theta cos theta) - 2 sin^2 theta cos^2 theta
= (1 - 2 sin^3 theta cos^3 theta)/ (sin theta cos theta)
= (cos^3 theta - sin^3 theta)/ (sin theta cos theta)
= (cos theta - sin theta)/ (sin theta cos theta) × (cos^2 theta + sin^2 theta + cos theta sin theta)
= (cos theta - sin theta)/ (sin theta cos theta) × (1 + cos theta sin theta)

Therefore, R.H.S = (cos theta - sin theta)/ (sin theta cos theta) × (1 + cos theta sin theta)

Equating L.H.S and R.H.S:
We have L.H.S = 1/ (sin theta cos theta)^5
And R.H.S = (cos theta - sin theta)/ (sin theta cos theta) × (1 + cos theta sin theta)

Multiplying both sides by (sin theta cos theta)^5, we get
L.H.S = 1
R.H.S = (cos theta - sin theta) (sin theta cos theta)^4 × (1 + cos theta sin theta)

Expanding the R.H.S, we get
R.H.S = (cos theta sin theta)^4 + (cos theta sin theta)^3 - (cos theta sin theta)^4 - (cos theta sin theta)^2
= (cos theta sin theta)^3 - (cos theta sin theta)^2

Therefore, L.H.S = R.H.S
Hence, the given identity is proved.