If a is an angle in first quadrant such that `"cosec"^(4) alpha = 17+"cot"^(4) alpha` ,then what is the value of `sin alpha` ?

Options

A. a)`1/3`
B. b)`1/4`
C. c)`1/9`
D. d)`1/16`

Answer

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

\[ \csc^4 \alpha = 17 + \cot^4 \alpha \]

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \csc^2 \alpha = 1 + \cot^2 \alpha \]
Thus, we can express \(\cot^4 \alpha\) in terms of \(\csc^2 \alpha\):
\[ \cot^4 \alpha = (\csc^2 \alpha - 1)^2 \]

### Step 2: Substitute \(\cot^4 \alpha\) in the original equation
Substituting \(\cot^4 \alpha\) into the original equation gives:
\[ \csc^4 \alpha = 17 + (\csc^2 \alpha - 1)^2 \]

### Step 3: Expand the equation
Expanding \((\csc^2 \alpha - 1)^2\):
\[ (\csc^2 \alpha - 1)^2 = \csc^4 \alpha - 2\csc^2 \alpha + 1 \]
Now substituting this back into the equation:
\[ \csc^4 \alpha = 17 + \csc^4 \alpha - 2\csc^2 \alpha + 1 \]

### Step 4: Simplify the equation
Subtract \(\csc^4 \alpha\) from both sides:
\[ 0 = 18 - 2\csc^2 \alpha \]
Rearranging gives:
\[ 2\csc^2 \alpha = 18 \]
Dividing both sides by 2:
\[ \csc^2 \alpha = 9 \]

### Step 5: Find \(\sin \alpha\)
Since \(\csc \alpha = \frac{1}{\sin \alpha}\), we have:
\[ \csc^2 \alpha = \frac{1}{\sin^2 \alpha} \]
Thus:
\[ \frac{1}{\sin^2 \alpha} = 9 \]
Taking the reciprocal:
\[ \sin^2 \alpha = \frac{1}{9} \]
Taking the square root (and considering that \(\alpha\) is in the first quadrant where sine is positive):
\[ \sin \alpha = \frac{1}{3} \]

### Final Answer
The value of \(\sin \alpha\) is:
\[ \sin \alpha = \frac{1}{3} \]

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