If `sin^(4)theta+cos^(4)theta=2sin^(2)thetacos^(2)theta,theta` is an acute angle , then the value of `tantheta` is

Options

A. 1
B. 2
C. `sqrt(2)`
D. 0

Answer

To solve the equation \( \sin^4 \theta + \cos^4 \theta = 2 \sin^2 \theta \cos^2 \theta \), we can start by rewriting the left-hand side using the identity for the sum of squares.

### Step 1: Rewrite the left-hand side
We know that:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we can substitute this into the equation:
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 2: Set the equation
Now we can set the equation:
\[ 1 - 2\sin^2 \theta \cos^2 \theta = 2\sin^2 \theta \cos^2 \theta \]

### Step 3: Combine like terms
Combine the terms involving \( \sin^2 \theta \cos^2 \theta \):
\[ 1 = 2\sin^2 \theta \cos^2 \theta + 2\sin^2 \theta \cos^2 \theta \]
This simplifies to:
\[ 1 = 4\sin^2 \theta \cos^2 \theta \]

### Step 4: Solve for \( \sin^2 \theta \cos^2 \theta \)
Now, we can solve for \( \sin^2 \theta \cos^2 \theta \):
\[ \sin^2 \theta \cos^2 \theta = \frac{1}{4} \]

### Step 5: Use the identity for tangent
We know that:
\[ \sin^2 \theta \cos^2 \theta = \frac{1}{4} \Rightarrow \frac{1}{4} = \sin^2 \theta \cdot (1 - \sin^2 \theta) \]
Let \( x = \sin^2 \theta \). Then we have:
\[ x(1 - x) = \frac{1}{4} \]
This expands to:
\[ x - x^2 = \frac{1}{4} \]
Rearranging gives us:
\[ x^2 - x + \frac{1}{4} = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{1 \pm \sqrt{1 - 1}}{2} = \frac{1}{2} \]
Thus, \( \sin^2 \theta = \frac{1}{2} \).

### Step 7: Find \( \tan \theta \)
Since \( \sin^2 \theta = \frac{1}{2} \), we have:
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{1}{2} = \frac{1}{2} \]
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{\frac{1}{2}}}{\sqrt{\frac{1}{2}}} = 1 \]

### Conclusion
Thus, the value of \( \tan \theta \) is:
\[ \boxed{1} \]