Question

If $ \tan\left( \alpha - \frac{\pi}{12} \right) = \frac{1}{\sqrt{3}} $, find $ \alpha $.

• A. \( \alpha = \frac{\pi}{6} \)
• B. \( \alpha = \frac{\pi}{4} \)
• C. \( \alpha = \frac{\pi}{3} \)
• D. \( \alpha = \frac{\pi}{2} \)

Hint:

When solving trigonometric equations, try to recall basic values of trigonometric functions like \( \sin, \cos, \tan \), especially for common angles like \( 30^\circ \), \( 45^\circ \), and \( 60^\circ \).

Answer 1

The Correct Option is A

We are given: \[ \tan \left( \alpha - \frac{\pi}{12} \right) = \frac{1}{\sqrt{3}} \] We know that: \[ \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \]
Thus, we can conclude: \[ \alpha - \frac{\pi}{12} = \frac{\pi}{6} \] Solving for \( \alpha \): \[ \alpha = \frac{\pi}{6} + \frac{\pi}{12} = \frac{2\pi}{12} + \frac{\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4} \]
Thus, \( \alpha = \frac{\pi}{6} \).