If `cosectheta+cosec^(2)theta=1` , then what is the value of `(cot^(12)theta-3cot^(10)theta+3cot^(8)theta-cot^(6)theta)`?

Options

A. `-2`
B. `-1`
C. 0
D. 1

Answer

To solve the equation \( \csc \theta + \csc^2 \theta = 1 \) and find the value of \( \cot^{12} \theta - 3 \cot^{10} \theta + 3 \cot^8 \theta - \cot^6 \theta \), we can follow these steps:

### Step 1: Rewrite the equation
The equation given is:
\[ \csc \theta + \csc^2 \theta = 1 \]
We know that \( \csc \theta = \frac{1}{\sin \theta} \) and \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \). Therefore, we can rewrite the equation as:
\[ \frac{1}{\sin \theta} + \frac{1}{\sin^2 \theta} = 1 \]

### Step 2: Multiply through by \( \sin^2 \theta \)
To eliminate the fractions, multiply the entire equation by \( \sin^2 \theta \):
\[ \sin^2 \theta \cdot \frac{1}{\sin \theta} + \sin^2 \theta \cdot \frac{1}{\sin^2 \theta} = \sin^2 \theta \]
This simplifies to:
\[ \sin \theta + 1 = \sin^2 \theta \]

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

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -1, c = -1 \):
\[ \sin \theta = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

### Step 5: Determine \( \cot^2 \theta \)
Using \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \cos^2 \theta \):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Calculating \( \sin^2 \theta \):
\[ \sin^2 \theta = \left(\frac{1 + \sqrt{5}}{2}\right)^2 = \frac{1 + 2\sqrt{5} + 5}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]
Thus,
\[ \cos^2 \theta = 1 - \frac{3 + \sqrt{5}}{2} = \frac{2 - (3 + \sqrt{5})}{2} = \frac{-1 - \sqrt{5}}{2} \]
Since \( \cos^2 \theta \) cannot be negative, we take the positive root for \( \sin \theta \):
\[ \sin \theta = \frac{1 + \sqrt{5}}{2} \]
Then we find \( \cot^2 \theta \):
\[ \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \]

### Step 6: Substitute into the polynomial
Let \( x = \cot^2 \theta \). The expression we need to evaluate becomes:
\[ x^6 - 3x^5 + 3x^4 - x^3 \]
This can be factored as:
\[ (x^2 - 1)^3 \]
Thus, we evaluate:
\[ (x^2 - 1)^3 = (0)^3 = 0 \]

### Final Answer
The value of \( \cot^{12} \theta - 3 \cot^{10} \theta + 3 \cot^8 \theta - \cot^6 \theta \) is:
\[ \boxed{0} \]