If `cos^(2)x+cos^(4)x=1`, then find the value of `tan^(2)x+tan^(4)x`.

Options

A. 0
B. 3
C. 1
D. 4

Answer

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

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^2 x + \cos^4 x = 1 \]

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

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

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

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

### Step 6: Find \( \tan^2 x \)
Recall that:
\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{1 - \cos^2 x}{\cos^2 x} \]
Substituting \( y \) into this gives:
\[ \tan^2 x = \frac{1 - y}{y} = \frac{1 - \frac{-1 + \sqrt{5}}{2}}{\frac{-1 + \sqrt{5}}{2}} \]
\[ = \frac{\frac{3 - \sqrt{5}}{2}}{\frac{-1 + \sqrt{5}}{2}} = \frac{3 - \sqrt{5}}{-1 + \sqrt{5}} \]

### Step 7: Find \( \tan^4 x \)
Now, we need \( \tan^4 x \):
\[ \tan^4 x = \left(\tan^2 x\right)^2 = \left(\frac{3 - \sqrt{5}}{-1 + \sqrt{5}}\right)^2 \]

### Step 8: Combine \( \tan^2 x + \tan^4 x \)
Now we can find \( \tan^2 x + \tan^4 x \):
\[ \tan^2 x + \tan^4 x = \tan^2 x \left(1 + \tan^2 x\right) \]
Using the identity \( 1 + \tan^2 x = \sec^2 x = \frac{1}{\cos^2 x} \):
\[ = \tan^2 x \cdot \frac{1}{y} \]

### Step 9: Substitute \( y \) back
Substituting \( y = \frac{-1 + \sqrt{5}}{2} \):
\[ \tan^2 x + \tan^4 x = \frac{(3 - \sqrt{5})}{(-1 + \sqrt{5})} \cdot \frac{2}{-1 + \sqrt{5}} \]

### Final Result
After simplifying, we find:
\[ \tan^2 x + \tan^4 x = 2 \]