Q.If \( \cos\theta - \sin\theta = \sqrt{2} \sin\theta \), then the value of \( \cos\theta + \sin\theta - \sqrt{2} \cos\theta \) is — (a) \(0\) (b) \(1\) (c) \(\sqrt2\) (d) None of the above
Answer: A
Given: \( \cos\theta - \sin\theta = \sqrt{2} \sin\theta \) ⇒ \( \sqrt{2} \sin\theta = \cos\theta - \sin\theta \) ⇒ \( (\sqrt{2} \sin\theta)^2 = (\cos\theta - \sin\theta)^2 \) ⇒ \( 2 \sin^2\theta = \cos^2\theta + \sin^2\theta - 2 \sin\theta \cos\theta \) ⇒ \( 2 \sin^2\theta - \sin^2\theta + 2 \sin\theta \cos\theta = \cos^2\theta \) ⇒ \( \sin^2\theta + 2 \sin\theta \cos\theta = \cos^2\theta \) ⇒ \( \cos^2\theta + \sin^2\theta + 2 \sin\theta \cos\theta = \cos^2\theta + \cos^2\theta \) ⇒ \( (\cos\theta + \sin\theta)^2 = 2 \cos^2\theta \) ⇒ \( \cos\theta + \sin\theta = \sqrt{2 \cos^2\theta} = \sqrt{2} \cos\theta \) ⇒ \( \cos\theta + \sin\theta - \sqrt{2} \cos\theta = 0 \)
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