Question

If \( \tan\theta+\sec\theta=\sqrt{3} \), then the principal value of \( \theta \) in \( [0,2\pi] \) is

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

Hint:

For equations involving \(\tan\theta+\sec\theta\), always check standard angles like \(30^\circ,45^\circ,60^\circ\).

Answer 1

The Correct Option is B

Concept: We use standard trigonometric values to solve equations involving \(\tan\theta\) and \(\sec\theta\).

Step 1: Given: \[ \tan\theta+\sec\theta=\sqrt{3} \]

Step 2: Check the standard angle \( \theta=\frac{\pi}{6} \).
We know: \[ \tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} \] and \[ \sec\frac{\pi}{6}=\frac{1}{\cos\frac{\pi}{6}} \] Since \[ \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \] therefore \[ \sec\frac{\pi}{6}=\frac{2}{\sqrt{3}} \]

Step 3: Add both values. \[ \tan\frac{\pi}{6}+\sec\frac{\pi}{6} = \frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}} \] \[ =\frac{3}{\sqrt{3}} \] \[ =\sqrt{3} \]

Step 4: Hence, the equation is satisfied by: \[ \theta=\frac{\pi}{6} \] Therefore, the principal value is: \[ \boxed{\frac{\pi}{6}} \]