If `tan theta + sec theta = sqrt3, 0 lt theta le pi`, then `theta =`

Options

A. `pi//3`
B. `2pi//3`
C. `pi//6`
D. `5pi//8`

Answer

To solve the equation \( \tan \theta + \sec \theta = \sqrt{3} \) for \( 0 < \theta \leq \pi \), we will follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \]
Substituting these into the equation gives:
\[ \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \sqrt{3} \]
Combining the fractions, we have:
\[ \frac{\sin \theta + 1}{\cos \theta} = \sqrt{3} \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ \sin \theta + 1 = \sqrt{3} \cos \theta \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[ \sin \theta = \sqrt{3} \cos \theta - 1 \]

### Step 4: Square both sides
Squaring both sides to eliminate the square root gives:
\[ \sin^2 \theta = (\sqrt{3} \cos \theta - 1)^2 \]
Expanding the right side:
\[ \sin^2 \theta = 3 \cos^2 \theta - 2\sqrt{3} \cos \theta + 1 \]

### Step 5: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can replace \( \sin^2 \theta \):
\[ 1 - \cos^2 \theta = 3 \cos^2 \theta - 2\sqrt{3} \cos \theta + 1 \]
Simplifying this gives:
\[ - \cos^2 \theta = 3 \cos^2 \theta - 2\sqrt{3} \cos \theta \]
Rearranging leads to:
\[ 0 = 4 \cos^2 \theta - 2\sqrt{3} \cos \theta \]

### Step 6: Factor the equation
Factoring out \( 2 \cos \theta \):
\[ 2 \cos \theta (2 \cos \theta - \sqrt{3}) = 0 \]

### Step 7: Solve for \( \cos \theta \)
Setting each factor to zero gives:
1. \( 2 \cos \theta = 0 \) → \( \cos \theta = 0 \)
2. \( 2 \cos \theta - \sqrt{3} = 0 \) → \( \cos \theta = \frac{\sqrt{3}}{2} \)

### Step 8: Find the corresponding angles
1. For \( \cos \theta = 0 \):
- \( \theta = \frac{\pi}{2} \)

2. For \( \cos \theta = \frac{\sqrt{3}}{2} \):
- \( \theta = \frac{\pi}{6} \) (in the first quadrant)
- \( \theta = \frac{11\pi}{6} \) (but this is outside the range \( 0 < \theta \leq \pi \))

### Step 9: Conclusion
Thus, the only valid solution in the given range is:
\[ \theta = \frac{\pi}{6} \]