If `sinx+cosecx=2," then "sin^nx+cosec^nx` is equal to

Options

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

Explanation

To solve the problem, we start with the given equation:

**Given:**
\[ \sin x + \csc x = 2 \]

### Step 1: Rewrite the equation
Recall that \(\csc x = \frac{1}{\sin x}\). Therefore, we can rewrite the equation as:
\[ \sin x + \frac{1}{\sin x} = 2 \]

### Step 2: Multiply through by \(\sin x\)
To eliminate the fraction, multiply both sides by \(\sin x\):
\[ \sin^2 x + 1 = 2 \sin x \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \sin^2 x - 2 \sin x + 1 = 0 \]

### Step 4: Factor the quadratic equation
This can be factored as:
\[ (\sin x - 1)^2 = 0 \]

### Step 5: Solve for \(\sin x\)
From the factored form, we find:
\[ \sin x - 1 = 0 \implies \sin x = 1 \]

### Step 6: Find \(\csc x\)
Since \(\sin x = 1\), we can find \(\csc x\):
\[ \csc x = \frac{1}{\sin x} = \frac{1}{1} = 1 \]

### Step 7: Calculate \(\sin^n x + \csc^n x\)
Now we need to calculate:
\[ \sin^n x + \csc^n x \]
Substituting the values we found:
\[ \sin^n x + \csc^n x = 1^n + 1^n = 1 + 1 = 2 \]

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

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