In `tantheta+sec theta=sqrt3,0ltthetaltpi, then theta` is equal to

Options

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

Answer

To solve the equation \( \tan \theta + \sec \theta = \sqrt{3} \) where \( 0 < \theta < \pi \), we can 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:
\[ \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \sqrt{3} \]

### Step 2: Combine the fractions
Taking the common denominator, we have:
\[ \frac{\sin \theta + 1}{\cos \theta} = \sqrt{3} \]

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

### Step 4: Rearrange the equation
Rearranging the equation results in:
\[ \sin \theta = \sqrt{3} \cos \theta - 1 \]

### Step 5: Square both sides
To eliminate the sine and cosine, we square both sides:
\[ \sin^2 \theta = (\sqrt{3} \cos \theta - 1)^2 \]
Expanding the right side:
\[ \sin^2 \theta = 3 \cos^2 \theta - 2\sqrt{3} \cos \theta + 1 \]

### Step 6: Use the Pythagorean identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute \( \sin^2 \theta \) with \( 1 - \cos^2 \theta \):
\[ 1 - \cos^2 \theta = 3 \cos^2 \theta - 2\sqrt{3} \cos \theta + 1 \]

### Step 7: Simplify the equation
Subtracting 1 from both sides:
\[ -\cos^2 \theta = 3 \cos^2 \theta - 2\sqrt{3} \cos \theta \]
Rearranging gives:
\[ 0 = 4 \cos^2 \theta - 2\sqrt{3} \cos \theta \]

### Step 8: Factor the equation
Factoring out \( 2 \cos \theta \):
\[ 2 \cos \theta (2 \cos \theta - \sqrt{3}) = 0 \]

### Step 9: Solve for \( \cos \theta \)
Setting each factor to zero gives:
1. \( 2 \cos \theta = 0 \) which implies \( \cos \theta = 0 \) (not valid in the given range)
2. \( 2 \cos \theta - \sqrt{3} = 0 \) which implies \( \cos \theta = \frac{\sqrt{3}}{2} \)

### Step 10: Find \( \theta \)
The solutions for \( \theta \) where \( \cos \theta = \frac{\sqrt{3}}{2} \) are:
\[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{11\pi}{6} \quad (\text{but } \frac{11\pi}{6} \text{ is not in the range } 0 < \theta < \pi) \]

Thus, the solution is:
\[ \theta = \frac{\pi}{6} \]

### Final Answer
The value of \( \theta \) is \( \frac{\pi}{6} \).

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