If `sin^(4)theta-cos^(4)=k^(4), then sin^(2)theta-cos^(2)theta` is __________.

Options

A. `K^(4)`
B. `K^(3)`
C. `K^(2)`
D. `K`

Explanation

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

\[ \sin^4 \theta - \cos^4 \theta = k^4 \]

We can use the difference of squares formula, which states that \(A^2 - B^2 = (A - B)(A + B)\). Here, we can let \(A = \sin^2 \theta\) and \(B = \cos^2 \theta\). Thus, we can rewrite the equation as:

\[ (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) = k^4 \]

Next, we know from the Pythagorean identity that:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Substituting this into our equation gives us:

\[ (\sin^2 \theta - \cos^2 \theta)(1) = k^4 \]

This simplifies to:

\[ \sin^2 \theta - \cos^2 \theta = k^4 \]

Thus, the value of \(\sin^2 \theta - \cos^2 \theta\) is:

\[ \sin^2 \theta - \cos^2 \theta = k^4 \]

So, the final answer is:

\[ \sin^2 \theta - \cos^2 \theta = k^4 \]