Solve for `theta(0^(@) lt theta lt 90^(@))` :
`tan^(2)theta+3=3sectheta`.

Answer

To solve the equation \( \tan^2 \theta + 3 = 3 \sec \theta \) for \( \theta \) in the interval \( 0^\circ < \theta < 90^\circ \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Substituting this into the equation gives:
\[ \sec^2 \theta - 1 + 3 = 3 \sec \theta \]

### Step 2: Simplify the equation
Now, simplify the equation:
\[ \sec^2 \theta + 2 = 3 \sec \theta \]
Rearranging gives:
\[ \sec^2 \theta - 3 \sec \theta + 2 = 0 \]

### Step 3: Factor the quadratic equation
This is a quadratic equation in terms of \( \sec \theta \). We can factor it:
\[ (\sec \theta - 1)(\sec \theta - 2) = 0 \]

### Step 4: Solve for \( \sec \theta \)
Setting each factor to zero gives us:
1. \( \sec \theta - 1 = 0 \) → \( \sec \theta = 1 \)
2. \( \sec \theta - 2 = 0 \) → \( \sec \theta = 2 \)

### Step 5: Find \( \theta \) from \( \sec \theta \)
Recall that \( \sec \theta = \frac{1}{\cos \theta} \):
1. For \( \sec \theta = 1 \):
\[ \cos \theta = 1 \quad \Rightarrow \quad \theta = 0^\circ \]
(This solution is not valid since \( \theta \) must be greater than \( 0^\circ \).)

2. For \( \sec \theta = 2 \):
\[ \cos \theta = \frac{1}{2} \quad \Rightarrow \quad \theta = 60^\circ \]

### Final Answer
Thus, the solution for \( \theta \) in the interval \( 0^\circ < \theta < 90^\circ \) is:
\[ \theta = 60^\circ \]

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