When `theta = 45 ^(@)` then what is the value of `(sec theta + cosec theta) ` ?

Options

A. `3 sqrt2`
B. `4 sqrt2`
C. `2 sqrt2`
D. `5 sqrt2`

Answer

To solve the problem, we need to find the value of \( \sec \theta + \csc \theta \) when \( \theta = 45^\circ \).

### Step-by-step Solution:

1. **Identify the Trigonometric Functions:**
- We need to calculate \( \sec \theta \) and \( \csc \theta \).
- Recall the definitions:
- \( \sec \theta = \frac{1}{\cos \theta} \)
- \( \csc \theta = \frac{1}{\sin \theta} \)

2. **Find the Values of \( \sin \theta \) and \( \cos \theta \) at \( \theta = 45^\circ \):**
- From trigonometric ratios, we know:
- \( \sin 45^\circ = \frac{\sqrt{2}}{2} \)
- \( \cos 45^\circ = \frac{\sqrt{2}}{2} \)

3. **Calculate \( \sec 45^\circ \) and \( \csc 45^\circ \):**
- Using the definitions:
- \( \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \)
- \( \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \)

4. **Add the Values:**
- Now, we add \( \sec 45^\circ \) and \( \csc 45^\circ \):
\[ \sec 45^\circ + \csc 45^\circ = \sqrt{2} + \sqrt{2} = 2\sqrt{2} \]

5. **Final Answer:**
- Therefore, the value of \( \sec \theta + \csc \theta \) when \( \theta = 45^\circ \) is \( 2\sqrt{2} \).