To solve for [tex]\(\tan(\theta)\)[/tex] given [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant IV, we can follow these steps:
1. Use the Pythagorean Identity:
The Pythagorean trigonometric identity states:
[tex]\[
\cos^2(\theta) + \sin^2(\theta) = 1
\][/tex]
Given [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex], substitute this into the identity:
[tex]\[
\left(\frac{8}{9}\right)^2 + \sin^2(\theta) = 1
\][/tex]
Calculate [tex]\(\left(\frac{8}{9}\right)^2\)[/tex]:
[tex]\[
\left(\frac{8}{9}\right)^2 = \frac{64}{81}
\][/tex]
Substitute back into the equation:
[tex]\[
\frac{64}{81} + \sin^2(\theta) = 1
\][/tex]
2. Solve for [tex]\(\sin^2(\theta)\)[/tex]:
Subtract [tex]\(\frac{64}{81}\)[/tex] from 1:
[tex]\[
\sin^2(\theta) = 1 - \frac{64}{81}
\][/tex]
Convert 1 to a fraction with a denominator of 81:
[tex]\[
1 = \frac{81}{81}
\][/tex]
Subtract the fractions:
[tex]\[
\sin^2(\theta) = \frac{81}{81} - \frac{64}{81} = \frac{17}{81}
\][/tex]
3. Find [tex]\(\sin(\theta)\)[/tex]:
Take the square root of both sides:
[tex]\[
\sin(\theta) = \pm \sqrt{\frac{17}{81}}
\][/tex]
Simplify the square root:
[tex]\[
\sin(\theta) = \pm \frac{\sqrt{17}}{9}
\][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant IV, [tex]\(\sin(\theta)\)[/tex] is negative:
[tex]\[
\sin(\theta) = -\frac{\sqrt{17}}{9}
\][/tex]
4. Calculate [tex]\(\tan(\theta)\)[/tex]:
The tangent function is given by:
[tex]\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\][/tex]
Substitute the values:
[tex]\[
\tan(\theta) = \frac{-\frac{\sqrt{17}}{9}}{\frac{8}{9}}
\][/tex]
Simplify the fraction:
[tex]\[
\tan(\theta) = -\frac{\sqrt{17}}{9} \times \frac{9}{8} = -\frac{\sqrt{17}}{8}
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{-\frac{\sqrt{17}}{8}}
\][/tex]
