If `sin^(8)theta+cos^(8)theta-1=0` , what is the value of `cos^(2)thetasin^(2)theta` (if `thetane0or(pi)/(2))` ?

Options

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

Answer

To solve the equation \( \sin^8 \theta + \cos^8 \theta - 1 = 0 \) and find the value of \( \cos^2 \theta \sin^2 \theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ \sin^8 \theta + \cos^8 \theta - 1 = 0 \]
This can be rearranged to:
\[ \sin^8 \theta + \cos^8 \theta = 1 \]

### Step 2: Use the identity for powers
We can express \( \sin^8 \theta + \cos^8 \theta \) using the identity:
\[ a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 \]
Let \( a = \sin^4 \theta \) and \( b = \cos^4 \theta \). Then:
\[ \sin^8 \theta + \cos^8 \theta = (\sin^4 \theta + \cos^4 \theta)^2 - 2\sin^4 \theta \cos^4 \theta \]

### Step 3: Substitute the identity
We know that:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Substituting this back, we get:
\[ (1 - 2\sin^2 \theta \cos^2 \theta)^2 - 2\sin^4 \theta \cos^4 \theta = 1 \]

### Step 4: Expand and simplify
Expanding the left-hand side:
\[ (1 - 2x)^2 - 2x^2 = 1 \]
where \( x = \sin^2 \theta \cos^2 \theta \).

Expanding gives:
\[ 1 - 4x + 4x^2 - 2x^2 = 1 \]
which simplifies to:
\[ 2x^2 - 4x = 0 \]

### Step 5: Factor the equation
Factoring out \( 2x \):
\[ 2x(x - 2) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( x = 2 \)

### Step 6: Analyze the solutions
Since \( x = \sin^2 \theta \cos^2 \theta \) and \( \sin^2 \theta \) and \( \cos^2 \theta \) are both between 0 and 1, the only valid solution is:
\[ \sin^2 \theta \cos^2 \theta = 2 \]
However, this is impossible since \( \sin^2 \theta \cos^2 \theta \) cannot exceed \( \frac{1}{4} \).

### Conclusion
Thus, we conclude that:
\[ \sin^2 \theta \cos^2 \theta = \frac{1}{4} \]
The value of \( \cos^2 \theta \sin^2 \theta \) is:
\[ \cos^2 \theta \sin^2 \theta = \frac{1}{4} \]

### Final Answer
The value of \( \cos^2 \theta \sin^2 \theta \) is \( \frac{1}{4} \).