If `sec theta + cos theta =(5)/(2)," where "O lt theta lt (pi)/(2)`, then find the value of `sec theta - cos theta =` ?
यदि `O lt theta lt (pi)/(2)` की स्थिति में `sec theta + cos theta =(5)/(2)` हो, तो s`sec theta - cos theta` का मान कितना होगा?

Options

A. `(3)/(2)`
B. `(2)/(3)`
C. `(1)/(2)`
D. `(3)/(4)`

Explanation

To solve the problem, we need to find the value of \( \sec \theta - \cos \theta \) given that \( \sec \theta + \cos \theta = \frac{5}{2} \) and \( 0 < \theta < \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sec \theta + \cos \theta = \frac{5}{2} \]

2. **Recall the relationship between secant and cosine:**
\[ \sec \theta = \frac{1}{\cos \theta} \]
Substitute this into the equation:
\[ \frac{1}{\cos \theta} + \cos \theta = \frac{5}{2} \]

3. **Multiply through by \( \cos \theta \) to eliminate the fraction:**
\[ 1 + \cos^2 \theta = \frac{5}{2} \cos \theta \]

4. **Rearrange the equation:**
\[ 2 + 2\cos^2 \theta = 5\cos \theta \]
\[ 2\cos^2 \theta - 5\cos \theta + 2 = 0 \]

5. **Let \( x = \cos \theta \) and rewrite the quadratic equation:**
\[ 2x^2 - 5x + 2 = 0 \]

6. **Use the quadratic formula to solve for \( x \):**
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]

7. **Calculate the two potential solutions:**
\[ x_1 = \frac{8}{4} = 2 \quad \text{(not valid since } \cos \theta \leq 1\text{)} \]
\[ x_2 = \frac{2}{4} = \frac{1}{2} \]

8. **Since \( \cos \theta = \frac{1}{2} \), find \( \sec \theta \):**
\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{1}{2}} = 2 \]

9. **Now calculate \( \sec \theta - \cos \theta \):**
\[ \sec \theta - \cos \theta = 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

### Final Answer:
\[ \sec \theta - \cos \theta = \frac{3}{2} \]