If `3 cos^2 theta - 4 sin theta + 1 = 0 ` and `0^@ lt theta lt 90^@` , then find the value of `sec theta + tan theta`

Options

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

Answer

To solve the equation \(3 \cos^2 \theta - 4 \sin \theta + 1 = 0\) and find the value of \( \sec \theta + \tan \theta\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 3 \cos^2 \theta - 4 \sin \theta + 1 = 0 \]
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta\), we can rewrite the equation in terms of \( \sin \theta\):
\[ 3(1 - \sin^2 \theta) - 4 \sin \theta + 1 = 0 \]
This simplifies to:
\[ 3 - 3 \sin^2 \theta - 4 \sin \theta + 1 = 0 \]
or:
\[ -3 \sin^2 \theta - 4 \sin \theta + 4 = 0 \]
Multiplying through by -1 gives:
\[ 3 \sin^2 \theta + 4 \sin \theta - 4 = 0 \]

### Step 2: Solve the quadratic equation
Now we can use the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3\), \( b = 4\), and \( c = -4\).
Calculating the discriminant:
\[ b^2 - 4ac = 4^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64 \]
Now substituting into the formula:
\[ \sin \theta = \frac{-4 \pm \sqrt{64}}{2 \cdot 3} = \frac{-4 \pm 8}{6} \]
Calculating the two possible values:
1. \( \sin \theta = \frac{4}{6} = \frac{2}{3} \)
2. \( \sin \theta = \frac{-12}{6} = -2 \) (not valid since \( \sin \theta \) must be between -1 and 1)

Thus, we have:
\[ \sin \theta = \frac{2}{3} \]

### Step 3: Find \cos \theta
Using the Pythagorean identity:
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{2}{3}\right)^2 = 1 - \frac{4}{9} = \frac{5}{9} \]
So:
\[ \cos \theta = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 4: Calculate \( \sec \theta + \tan \theta \)
Now we can find \( \sec \theta \) and \( \tan \theta \):
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{3}{\sqrt{5}} \]
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \]
Now adding them together:
\[ \sec \theta + \tan \theta = \frac{3}{\sqrt{5}} + \frac{2}{\sqrt{5}} = \frac{3 + 2}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5} \]

### Final Answer
Thus, the value of \( \sec \theta + \tan \theta \) is:
\[ \sqrt{5} \]