Solve `: tan theta + sec theta = sqrt3.`

Answer

To solve the equation \( \tan \theta + \sec \theta = \sqrt{3} \), we can follow these steps:

### Step 1: Isolate \( \sec \theta \)
We start by isolating \( \sec \theta \):
\[ \sec \theta = \sqrt{3} - \tan \theta \]

**Hint:** Rearranging the equation can help simplify the problem.

### Step 2: Square both sides
Next, we square both sides of the equation:
\[ \sec^2 \theta = (\sqrt{3} - \tan \theta)^2 \]
Expanding the right-hand side:
\[ \sec^2 \theta = 3 - 2\sqrt{3} \tan \theta + \tan^2 \theta \]

**Hint:** Remember to apply the square of a binomial formula: \( (a - b)^2 = a^2 - 2ab + b^2 \).

### Step 3: Substitute \( \sec^2 \theta \)
We know that \( \sec^2 \theta = 1 + \tan^2 \theta \). Substituting this into the equation gives:
\[ 1 + \tan^2 \theta = 3 - 2\sqrt{3} \tan \theta + \tan^2 \theta \]

**Hint:** Use trigonometric identities to replace \( \sec^2 \theta \).

### Step 4: Simplify the equation
Now, we can simplify the equation by canceling \( \tan^2 \theta \) from both sides:
\[ 1 = 3 - 2\sqrt{3} \tan \theta \]
Rearranging gives:
\[ 2\sqrt{3} \tan \theta = 3 - 1 \]
\[ 2\sqrt{3} \tan \theta = 2 \]

**Hint:** Isolate the term with \( \tan \theta \) to solve for it.

### Step 5: Solve for \( \tan \theta \)
Dividing both sides by \( 2\sqrt{3} \):
\[ \tan \theta = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \]

**Hint:** Simplifying fractions can help find the value of \( \tan \theta \).

### Step 6: Find \( \theta \)
Now, we need to find the angle \( \theta \) for which \( \tan \theta = \frac{1}{\sqrt{3}} \). The standard angle that satisfies this is:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

**Hint:** Remember that the tangent function is periodic, so consider all possible angles.

### Final Solution
Thus, the solution to the equation \( \tan \theta + \sec \theta = \sqrt{3} \) is:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]