Q.If \( \cos\theta - \sin\theta = \sqrt{2} \sin\theta \), then what is the value of \( \cos\theta + \sin\theta \)? (a) \(2 cos\theta\) (b) \(\sqrt2 sin \theta\) (c) \(2 sin\theta\) (d) \(\sqrt2 cos \theta\)
Answer: D
Given: \( \cos\theta - \sin\theta = \sqrt{2} \sin\theta \) Or, \( \sqrt{2} \sin\theta = \cos\theta - \sin\theta \) Then, squaring both sides: \( (\sqrt{2} \sin\theta)^2 = (\cos\theta - \sin\theta)^2 \) So, \( 2 \sin^2\theta = \cos^2\theta + \sin^2\theta - 2 \sin\theta \cos\theta \) Rearranging: \( 2 \sin^2\theta - \sin^2\theta + 2 \sin\theta \cos\theta = \cos^2\theta \) Which simplifies to: \( \sin^2\theta + 2 \sin\theta \cos\theta = \cos^2\theta \) Now, adding \( \cos^2\theta \) to both sides: \( \cos^2\theta + \sin^2\theta + 2 \sin\theta \cos\theta = \cos^2\theta + \cos^2\theta \) So, \( (\cos\theta + \sin\theta)^2 = 2 \cos^2\theta \) Taking square root: \( \cos\theta + \sin\theta = \sqrt{2 \cos^2\theta} = \sqrt{2} \cos\theta \)
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