Given: $\sin\theta + \cos\theta = \sqrt{3}$
Squaring both sides:
$$(\sin\theta + \cos\theta)^2 = (\sqrt{3})^2$$
$$\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 3$$
Since $\sin^2\theta + \cos^2\theta = 1$:
$$1 + 2\sin\theta\cos\theta = 3$$
$$2\sin\theta\cos\theta = 2$$
$$\therefore \sin\theta\cdot\cos\theta = 1$$
The key trick is to square both sides of the given equation, then apply the identity $\sin^2\theta + \cos^2\theta = 1$. This converts the sum into a product directly. Examiners award 1 mark for the correct squaring step and 1 mark for the final value. Don't forget to substitute the Pythagorean identity before solving.