If `cosec^(4)alpha=17+cot^(4)alpha`, then find `sinalpha`.

Answer

To solve the equation \( \csc^4 \alpha = 17 + \cot^4 \alpha \) and find \( \sin \alpha \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \csc^4 \alpha = 17 + \cot^4 \alpha \]
We can express \( \csc^4 \alpha \) in terms of \( \cot^4 \alpha \):
\[ \csc^4 \alpha - \cot^4 \alpha = 17 \]

### Step 2: Use the identity for difference of squares
Recall the identity \( a^2 - b^2 = (a - b)(a + b) \). Here, let \( a = \csc^2 \alpha \) and \( b = \cot^2 \alpha \):
\[ (\csc^2 \alpha - \cot^2 \alpha)(\csc^2 \alpha + \cot^2 \alpha) = 17 \]

### Step 3: Use the Pythagorean identity
We know that:
\[ \csc^2 \alpha = 1 + \cot^2 \alpha \]
Substituting this into our equation:
\[ \csc^2 \alpha - \cot^2 \alpha = 1 \]
Thus, we have:
\[ (1)(\csc^2 \alpha + \cot^2 \alpha) = 17 \]
This simplifies to:
\[ \csc^2 \alpha + \cot^2 \alpha = 17 \]

### Step 4: Substitute \( \cot^2 \alpha \)
From the identity \( \csc^2 \alpha = 1 + \cot^2 \alpha \), we can substitute \( \cot^2 \alpha \) in terms of \( \csc^2 \alpha \):
\[ \csc^2 \alpha + (\csc^2 \alpha - 1) = 17 \]
This simplifies to:
\[ 2\csc^2 \alpha - 1 = 17 \]

### Step 5: Solve for \( \csc^2 \alpha \)
Adding 1 to both sides:
\[ 2\csc^2 \alpha = 18 \]
Dividing by 2:
\[ \csc^2 \alpha = 9 \]

### Step 6: Find \( \sin \alpha \)
Since \( \csc \alpha = \frac{1}{\sin \alpha} \), we have:
\[ \csc^2 \alpha = \frac{1}{\sin^2 \alpha} = 9 \]
Taking the reciprocal gives:
\[ \sin^2 \alpha = \frac{1}{9} \]
Taking the square root:
\[ \sin \alpha = \pm \frac{1}{3} \]

### Final Answer
Thus, the values of \( \sin \alpha \) are:
\[ \sin \alpha = \frac{1}{3} \quad \text{or} \quad \sin \alpha = -\frac{1}{3} \]