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