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

Step 1: Understanding the Question:
This problem asks for the specific angle \( \theta \) in the primary circular range that satisfies the equation \( \tan \theta + \sec \theta = \sqrt{3} \).
Step 2: Key Formula or Approach:
Method 1: Use the identity \( \sec^2 \theta - \tan^2 \theta = 1 \).
Method 2: Convert everything to \( \sin \) and \( \cos \).
Step 3: Detailed Explanation:

Method 1: We have \( \sec \theta + \tan \theta = \sqrt{3} \).
We know \( (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \).
Therefore, \( \sec \theta - \tan \theta = \frac{1}{\sqrt{3}} \).

Now, add the two equations:
\[ 2 \sec \theta = \sqrt{3} + \frac{1}{\sqrt{3}} \]
\[ 2 \sec \theta = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]
\[ \sec \theta = \frac{2}{\sqrt{3}} \implies \cos \theta = \frac{\sqrt{3}}{2} \]

Next, subtract the two equations:
\[ 2 \tan \theta = \sqrt{3} - \frac{1}{\sqrt{3}} \]
\[ 2 \tan \theta = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

From \( \cos \theta = \frac{\sqrt{3}}{2} \) and \( \tan \theta = \frac{1}{\sqrt{3}} \), the angle in the first quadrant that satisfies both is \( \theta = 30^\circ \) or \( \pi/6 \).

Step 4: Final Answer:
The principal value is \( \pi/6 \).