# ISI2015-DCG-64

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If $\cos 2 \theta = \sqrt{2}(\cos \theta – \sin \theta)$ 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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$cos 2\theta=cos^2\theta-sin^2\theta=(cos\theta+sin\theta)(cos\theta-sin\theta)$

$cos2\theta=\sqrt2(cos\theta-sin\theta)$

$\implies(cos\theta+sin\theta=\sqrt2)$

$\implies \theta=45°$

$\implies tan \theta=1$
Answer: $A$

Given:

$$\cos2\theta = \sqrt2(\cos\theta-\sin\theta) \qquad \to (1)$$

Now, we know that:

$$\cos2\theta = \cos^2\theta-\sin^2\theta = (\cos\theta + \sin\theta)(\cos\theta-\sin\theta)\qquad \to (2)$$

On comparing $(1)$ and $(2)$, we get:

$$\cos\theta + \sin\theta = \sqrt 2$$

Now, this is possible only when, $\theta = \frac{\pi}{4} = 45^0$

$$\implies \tan\theta = \tan\frac{\pi}{4} = 1$$

$\therefore A$ is the correct option.

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