If sin `theta`+ cosec `theta` =2, then `sin^n` `theta+"cosec"^n theta` is equal to

Options

A. 2
B. 2n
C. `2^(n-1)`
D. `2^(n-2)`

Answer

To solve the problem, we need to find the value of \( \sin^n \theta + \csc^n \theta \) given that \( \sin \theta + \csc \theta = 2 \).

### Step-by-Step Solution:

1. **Understanding the Given Equation:**
We start with the equation:
\[ \sin \theta + \csc \theta = 2 \]
Here, \( \csc \theta \) is the reciprocal of \( \sin \theta \), so we can rewrite it as:
\[ \sin \theta + \frac{1}{\sin \theta} = 2 \]

2. **Letting \( x = \sin \theta \):**
We can substitute \( x \) for \( \sin \theta \):
\[ x + \frac{1}{x} = 2 \]

3. **Multiplying through by \( x \):**
To eliminate the fraction, multiply both sides by \( x \):
\[ x^2 + 1 = 2x \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ x^2 - 2x + 1 = 0 \]

5. **Factoring the Quadratic:**
This can be factored as:
\[ (x - 1)^2 = 0 \]

6. **Finding the Value of \( x \):**
Solving for \( x \) gives:
\[ x - 1 = 0 \implies x = 1 \]
Therefore, \( \sin \theta = 1 \).

7. **Finding \( \csc \theta \):**
Since \( \sin \theta = 1 \), we have:
\[ \csc \theta = \frac{1}{\sin \theta} = 1 \]

8. **Calculating \( \sin^n \theta + \csc^n \theta \):**
Now we can find \( \sin^n \theta + \csc^n \theta \):
\[ \sin^n \theta + \csc^n \theta = 1^n + 1^n = 1 + 1 = 2 \]

### Final Answer:
Thus, the value of \( \sin^n \theta + \csc^n \theta \) is:
\[ \boxed{2} \]