To find \( \tan^2(90^\circ - x) \), we can use the co-function identity for tangent:
\[ \tan(90^\circ - x) = \cot(x) \]
Thus, we can express \( \tan^2(90^\circ - x) \) as:
\[ \tan^2(90^\circ - x) = \cot^2(x) = \frac{1}{\tan^2(x)} \]
Next, we need to find \( \tan(x) \) using the given value of \( \sin(x) = \frac{2}{3} \). We can utilize the Pythagorean identity:
\[ \sin^2(x) + \cos^2(x) = 1 \]
Calculating \( \cos^2(x) \):
\[ \sin^2(x) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \] \[ \cos^2(x) = 1 - \sin^2(x) = 1 - \frac{4}{9} = \frac{5}{9} \]
Now that we have \( \cos^2(x) \), we can find \( \tan(x) \):
\[ \tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\frac{2}{3}}{\sqrt{\frac{5}{9}}} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \]
Now we can calculate \( \tan^2(x) \):
\[ \tan^2(x) = \left(\frac{2}{\sqrt{5}}\right)^2 = \frac{4}{5} \]
Now, substituting this into our expression for \( \tan^2(90^\circ - x) \):
\[ \tan^2(90^\circ - x) = \frac{1}{\tan^2(x)} = \frac{1}{\frac{4}{5}} = \frac{5}{4} \]
Therefore, the final answer is:
\[ \tan^2(90^\circ - x) = \frac{5}{4} \]