Question

If \( \sin \theta + \cos \theta = \sqrt{2} \cos \theta \), then the value of \( \tan \theta \) is

• A. \( \sqrt{2} + 1 \)
• B. \( \sqrt{2} - 1 \)
• C. \( 1 - \sqrt{2} \)
• D. \( \frac{1}{\sqrt{2} + 1}

Hint:

\textbf{Key Fact:} Simplify trigonometric equations by isolating terms and converting to tangent for easier computation.

Answer 1

The Correct Option is B

• Simplify the Equation: Given \( \sin \theta + \cos \theta = \sqrt{2} \cos \theta \), rewrite as \[ \sin \theta + \cos \theta - \sqrt{2} \cos \theta = 0 \implies \sin \theta = (\sqrt{2} - 1) \cos \theta. \]
• Find \( \tan \theta \): Divide both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)): \[ \frac{\sin \theta}{\cos \theta} = \sqrt{2} - 1 \implies \tan \theta = \sqrt{2} - 1. \]
• Verification: Let \( \cos \theta = t \), so \( \sin \theta = (\sqrt{2} - 1) t \). Check if \[ \sin^2 \theta + \cos^2 \theta = ((\sqrt{2} - 1)^2 + 1) t^2 = 1. \] Since \[ (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}, \] the sum is \[ (3 - 2\sqrt{2}) + 1 = 4 - 2\sqrt{2}, \] so \[ t^2 = \frac{1}{4 - 2\sqrt{2}}, \] which is valid, confirming the solution.
• Conclusion: The correct value of \( \tan \theta \) is \( \boxed{\sqrt{2} - 1} \), corresponding to option (2).