Answer: $\mathbf A$
Given:
$$\cos2\theta = \sqrt2(\cos\theta-\sin\theta) \tag{1}$$
Now, we know that:
$$\cos2\theta = \cos^2\theta-\sin^2\theta = (\cos\theta + \sin\theta)(\cos\theta-\sin\theta) \tag{2}$$
On comparing $(1)$ and $(2)$, we get:
$$\cos\theta + \sin\theta = \sqrt 2$$
Now, this is possible only when, $\theta = \frac{\pi}{4} = 45^0$
$$\implies \tan\theta = \tan\frac{\pi}{4} = 1$$
$\therefore \mathbf A$ is the correct option.