If `tan^(2)theta - 3sectheta + 3=0, 0^(@) lt theta lt 90^(@)`, then the value of `sintheta + cos theta` is:

Options

A. A) `3sqrt(3)`
B. B) `(5sqrt(3))/3`
C. C) `2sqrt(3)`
D. D) `(5sqrt(3))/6`

Answer

To solve the equation \( \tan^2 \theta - 3 \sec \theta + 3 = 0 \) for \( 0^\circ < \theta < 90^\circ \) and find the value of \( \sin \theta + \cos \theta \), we can follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \]
Thus, we can rewrite \( \tan^2 \theta \) and \( \sec \theta \):
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Substituting this into the equation gives:
\[ \frac{\sin^2 \theta}{\cos^2 \theta} - 3 \cdot \frac{1}{\cos \theta} + 3 = 0 \]

### Step 2: Multiply through by \( \cos^2 \theta \) to eliminate the denominator
Multiplying the entire equation by \( \cos^2 \theta \) (noting that \( \cos \theta \neq 0 \) in the given range):
\[ \sin^2 \theta - 3 \cos \theta + 3 \cos^2 \theta = 0 \]

### Step 3: Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \)
From the identity, we can express \( \sin^2 \theta \) as:
\[ \sin^2 \theta = 1 - \cos^2 \theta \]
Substituting this into the equation gives:
\[ 1 - \cos^2 \theta - 3 \cos \theta + 3 \cos^2 \theta = 0 \]
This simplifies to:
\[ (1 + 2 \cos^2 \theta - 3 \cos \theta) = 0 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2 \cos^2 \theta - 3 \cos \theta + 1 = 0 \]

### Step 5: Solve the quadratic equation
Let \( x = \cos \theta \). The equation becomes:
\[ 2x^2 - 3x + 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
This gives us:
\[ x = \frac{4}{4} = 1 \quad \text{or} \quad x = \frac{2}{4} = \frac{1}{2} \]

### Step 6: Determine the values of \( \theta \)
Since \( \cos \theta = 1 \) corresponds to \( \theta = 0^\circ \) (not in the range), we take \( \cos \theta = \frac{1}{2} \), which corresponds to:
\[ \theta = 60^\circ \]

### Step 7: Calculate \( \sin \theta + \cos \theta \)
Now we find:
\[ \sin 60^\circ + \cos 60^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \]

### Final Answer
Thus, the value of \( \sin \theta + \cos \theta \) is:
\[ \frac{\sqrt{3} + 1}{2} \]