If `theta` be a positive acute angle satisfying `cos^(2)theta+cos^(4)theta=1` , then the value of `tan^(2)theta+tan^(4)theta` is

Options

A. `(3)/(2)`
B. 1
C. `(1)/(2)`
D. 0

Answer

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

1. **Given Equation**:
\[ \cos^2 \theta + \cos^4 \theta = 1 \]

2. **Substituting**:
Let \( x = \cos^2 \theta \). Then, we can rewrite the equation as:
\[ x + x^2 = 1 \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ x^2 + x - 1 = 0 \]

4. **Using the Quadratic Formula**:
To solve for \( x \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -1 \). Plugging in these values:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ x = \frac{-1 \pm \sqrt{5}}{2} \]

5. **Finding Positive Value**:
Since \( \theta \) is a positive acute angle, we take the positive root:
\[ x = \frac{-1 + \sqrt{5}}{2} \]
Thus,
\[ \cos^2 \theta = \frac{-1 + \sqrt{5}}{2} \]

6. **Finding \( \tan^2 \theta \)**:
We know that:
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \]
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[ \sin^2 \theta = 1 - \frac{-1 + \sqrt{5}}{2} = \frac{3 - \sqrt{5}}{2} \]
Therefore,
\[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\frac{3 - \sqrt{5}}{2}}{\frac{-1 + \sqrt{5}}{2}} = \frac{3 - \sqrt{5}}{-1 + \sqrt{5}} \]

7. **Simplifying \( \tan^2 \theta \)**:
To simplify \( \tan^2 \theta \):
\[ \tan^2 \theta = \frac{(3 - \sqrt{5})(-1 - \sqrt{5})}{(-1 + \sqrt{5})(-1 - \sqrt{5})} \]
The denominator becomes:
\[ (-1 + \sqrt{5})(-1 - \sqrt{5}) = 1 - 5 = -4 \]
The numerator simplifies to:
\[ (3 - \sqrt{5})(-1 - \sqrt{5}) = -3 - 3\sqrt{5} + \sqrt{5} + 5 = 2 - 2\sqrt{5} \]
Thus,
\[ \tan^2 \theta = \frac{2 - 2\sqrt{5}}{-4} = \frac{1 - \sqrt{5}}{-2} = \frac{\sqrt{5} - 1}{2} \]

8. **Finding \( \tan^4 \theta \)**:
Now, we calculate \( \tan^4 \theta \):
\[ \tan^4 \theta = \left(\tan^2 \theta\right)^2 = \left(\frac{\sqrt{5} - 1}{2}\right)^2 = \frac{(\sqrt{5} - 1)^2}{4} = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]

9. **Final Calculation**:
Now we add \( \tan^2 \theta \) and \( \tan^4 \theta \):
\[ \tan^2 \theta + \tan^4 \theta = \frac{\sqrt{5} - 1}{2} + \frac{3 - \sqrt{5}}{2} = \frac{\sqrt{5} - 1 + 3 - \sqrt{5}}{2} = \frac{2}{2} = 1 \]

Thus, the value of \( \tan^2 \theta + \tan^4 \theta \) is \( \boxed{1} \).