We are given the trigonometric equation and tasked to determine the correctness of Keisha's and David's procedures. Let's analyze the equation in step-by-step detail.
First, consider the original equation:
[tex]\[
\frac{\left(-\frac{8}{17}\right)^2}{\cos ^2 \theta} + 1 = \frac{1}{\cos ^2 \theta}
\][/tex]
1. Simplification of the given equation:
Simplify the left-hand side:
[tex]\[
\left(-\frac{8}{17}\right)^2 = \frac{64}{289}
\][/tex]
Thus, the equation becomes:
[tex]\[
\frac{\frac{64}{289}}{\cos^2 \theta} + 1 = \frac{1}{\cos^2 \theta}
\][/tex]
2. Eliminate the denominator:
Multiply through by [tex]\(\cos^2 \theta\)[/tex] to eliminate the denominator:
[tex]\[
\frac{64}{289} + \cos^2 \theta = 1
\][/tex]
3. Isolate [tex]\(\cos^2 \theta\)[/tex]:
Rearrange the equation to solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = 1 - \frac{64}{289}
\][/tex]
4. Simplify the right-hand side:
Simplify the fraction:
[tex]\[
\cos^2 \theta = \frac{289}{289} - \frac{64}{289} = \frac{225}{289}
\][/tex]
5. Take the square root:
Solve for [tex]\(\cos \theta\)[/tex] by taking the square root of both sides:
[tex]\[
\cos \theta = \pm \sqrt{\frac{225}{289}} = \pm \frac{15}{17}
\][/tex]
Now let's verify Keisha's and David's procedures:
- Keisha's procedure:
She solved for [tex]\(\cos \theta\)[/tex] as follows:
[tex]\[
\cos \theta = \pm \sqrt{1 - \frac{64}{289}} = \pm \sqrt{\frac{225}{289}} = \pm \frac{15}{17}
\][/tex]
This is correct.
- David's procedure:
He solved for [tex]\(\cos \theta\)[/tex] as follows:
[tex]\[
\cos \theta = \pm \sqrt{\frac{225}{289}} = \pm \frac{15}{17}
\][/tex]
This is also correct.
Therefore, both Keisha's and David's procedures are correct.
The final answer is:
[tex]\[
\text{Both procedures are correct.}
\][/tex]
