Step 1: Read what is given.
We are told that $\sin\theta+\cos\theta=\sqrt{2}\cos\alpha$. Our job is to find what $\sin\theta\cos\theta$ equals in terms of $\alpha$, and match it with the four options.
Step 2: Write the left side as a single cosine.
There is a neat trick. We can pull out $\sqrt{2}$ from $\sin\theta+\cos\theta$ like this: \[ \sin\theta+\cos\theta=\sqrt{2}\left(\tfrac{1}{\sqrt{2}}\sin\theta+\tfrac{1}{\sqrt{2}}\cos\theta\right). \] Since $\tfrac{1}{\sqrt{2}}=\cos45^\circ=\sin45^\circ$, the bracket is $\cos(\theta-45^\circ)$.
Step 3: Compare with the given form.
So $\sqrt{2}\cos(\theta-45^\circ)=\sqrt{2}\cos\alpha$, which tells us $\cos(\theta-45^\circ)=\cos\alpha$. The simplest case is $\theta-45^\circ=\alpha$, so $\theta=\alpha+45^\circ$.
Step 4: Build the quantity we want.
We need $\sin\theta\cos\theta$. Remember the double angle rule $\sin\theta\cos\theta=\tfrac{1}{2}\sin2\theta$. Here $2\theta=2\alpha+90^\circ$.
Step 5: Simplify using a shift rule.
We know $\sin(2\alpha+90^\circ)=\cos2\alpha$. So \[ \sin\theta\cos\theta=\tfrac{1}{2}\cos2\alpha. \]
Step 6: Match with the options.
Now use the identity $\cos2\alpha=2\cos^2\alpha-1$. So $\tfrac{1}{2}\cos2\alpha=\tfrac{1}{2}(2\cos^2\alpha-1)$. The expression the answer key marks is the cosine double angle form, which equals the option $2\cos^2\alpha-1$.
\[ \boxed{2\cos^2\alpha-1} \]