If `sin^(4)x + sin^(2) x =1`, then the value of `cot^(4)x + cot^(2)x` is:

Options

A. `cos^(2) x`
B. `sin^(2)x`
C. `tan^(2)x`
D. 1

Answer

To solve the equation \( \sin^4 x + \sin^2 x = 1 \) and find the value of \( \cot^4 x + \cot^2 x \), we can follow these steps:

### Step 1: Substitute \( y = \sin^2 x \)
Let \( y = \sin^2 x \). Then the equation becomes:
\[ y^2 + y = 1 \]

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

### Step 3: Solve the quadratic equation
We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \):
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 4: Find the positive root
Since \( y = \sin^2 x \) must be non-negative, we take the positive root:
\[ y = \frac{-1 + \sqrt{5}}{2} \]

### Step 5: Find \( \cot^2 x \)
Recall that \( \cot^2 x = \frac{\cos^2 x}{\sin^2 x} = \frac{1 - \sin^2 x}{\sin^2 x} \):
\[ \cot^2 x = \frac{1 - y}{y} = \frac{1 - \frac{-1 + \sqrt{5}}{2}}{\frac{-1 + \sqrt{5}}{2}} = \frac{\frac{3 - \sqrt{5}}{2}}{\frac{-1 + \sqrt{5}}{2}} = \frac{3 - \sqrt{5}}{-1 + \sqrt{5}} \]

### Step 6: Simplify \( \cot^2 x \)
To simplify \( \cot^2 x \), we multiply the numerator and denominator by the conjugate of the denominator:
\[ \cot^2 x = \frac{(3 - \sqrt{5})(-1 - \sqrt{5})}{(-1 + \sqrt{5})(-1 - \sqrt{5})} = \frac{(-3 - 3\sqrt{5} + \sqrt{5} + 5)}{1 - 5} = \frac{2 - 2\sqrt{5}}{-4} = \frac{1 - \sqrt{5}}{-2} \]

### Step 7: Find \( \cot^4 x + \cot^2 x \)
Now, we need to find \( \cot^4 x + \cot^2 x \):
Let \( z = \cot^2 x \), then:
\[ z^2 + z = \left(\frac{1 - \sqrt{5}}{-2}\right)^2 + \frac{1 - \sqrt{5}}{-2} \]

### Step 8: Substitute and simplify
Calculating \( z^2 \):
\[ z^2 = \left(\frac{1 - \sqrt{5}}{-2}\right)^2 = \frac{(1 - \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Now adding \( z \):
\[ z^2 + z = \frac{3 - \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{-2} = \frac{3 - \sqrt{5} + 1 - \sqrt{5}}{-2} = \frac{4 - 2\sqrt{5}}{-2} = -2 + \sqrt{5} \]

### Final Result
Thus, the value of \( \cot^4 x + \cot^2 x \) is:
\[ \cot^4 x + \cot^2 x = -2 + \sqrt{5} \]