If ${\sin }^2\theta -{\cos }^2\theta =-\frac{1}{3}$, find the value of ${\cos }^2\theta$.

The correct answer is $\frac{2}{3}$.

Step 1: Understand the given trigonometric equation
We are given the equation:

${\sin }^2\theta -{\cos }^2\theta =-\frac{1}{3}$

Step 2: Recall the fundamental trigonometric identity
We know that for any angle θ, the standard Pythagorean trigonometric identity is:

${\sin }^2\theta +{\cos }^2\theta =1$

From this identity, we can express ${\sin }^2\theta$ in terms of ${\cos }^2\theta$:

${\sin }^2\theta =1-{\cos }^2\theta$

Step 3: Substitute ${\sin }^2\theta$ into the given equation
Replace ${\sin }^2\theta$ with $1-{\cos }^2\theta$ in the original equation:

$(1-{\cos }^2\theta )-{\cos }^2\theta =-\frac{1}{3}$

Simplify the left side by combining like terms:

$1-2{\cos }^2\theta =-\frac{1}{3}$

Step 4: Solve for ${\cos }^2\theta$
Subtract 1 from both sides of the equation:

$-2{\cos }^2\theta =-\frac{1}{3}-1$

$-2{\cos }^2\theta =-\frac{4}{3}$

Divide both sides by -2:

${\cos }^2\theta =\frac{-4/3}{-2}$

${\cos }^2\theta =\frac{2}{3}$

Thus, the value of ${\cos }^2\theta$ is $\frac{2}{3}$.