If `sin x + cosec x =2`, then `sin^(3n) x + cosec ^(3n) x` equals

Options

A. `2^(2n-1)`
B. `2^(2n)`
C. `2^(3n+1)`
D. 2

Answer

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

**Step 1: Start with the equation**
\[ \sin x + \csc x = 2 \]

**Step 2: Rewrite \(\csc x\) in terms of \(\sin x\)**
\[ \sin x + \frac{1}{\sin x} = 2 \]

**Step 3: Multiply through by \(\sin x\) to eliminate the fraction**
\[ \sin^2 x + 1 = 2 \sin x \]

**Step 4: Rearrange the equation**
\[ \sin^2 x - 2 \sin x + 1 = 0 \]

**Step 5: Factor the quadratic equation**
\[ (\sin x - 1)^2 = 0 \]

**Step 6: Solve for \(\sin x\)**
\[ \sin x - 1 = 0 \implies \sin x = 1 \]

**Step 7: Find \(\csc x\) using \(\sin x\)**
Since \(\sin x = 1\),
\[ \csc x = \frac{1}{\sin x} = 1 \]

**Step 8: Substitute \(\sin x\) and \(\csc x\) into the expression**
Now we need to find \(\sin^{3n} x + \csc^{3n} x\):
\[ \sin^{3n} x + \csc^{3n} x = 1^{3n} + 1^{3n} = 1 + 1 = 2 \]

**Final Answer:**
\[ \sin^{3n} x + \csc^{3n} x = 2 \]