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If $\cos2\theta=\sqrt{2}(\cos\theta-\sin\theta)$ then $\tan\theta$ equals

  1. $1$
  2. $1$ or $-1$
  3. $\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$  or $1$
  4. None of these
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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.

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