Answer :
To determine the value of [tex]\(\tan(\theta)\)[/tex] given that [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant IV, we need to follow a systematic approach:
1. Identify the cosine value:
[tex]\[ \cos(\theta) = \frac{8}{9} \][/tex]
2. Determine the sine value using the Pythagorean identity:
The Pythagorean identity states:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Substitute [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex] into the identity:
[tex]\[ \sin^2(\theta) + \left(\frac{8}{9}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(\theta) + \frac{64}{81} = 1 \][/tex]
[tex]\[ \sin^2(\theta) = 1 - \frac{64}{81} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{81}{81} - \frac{64}{81} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{17}{81} \][/tex]
3. Solve for [tex]\(\sin(\theta)\)[/tex]:
Since [tex]\(\sin(\theta)\)[/tex] can be positive or negative:
[tex]\[ \sin(\theta) = \pm \sqrt{\frac{17}{81}} = \pm \frac{\sqrt{17}}{9} \][/tex]
We must consider the quadrant of [tex]\(\theta\)[/tex]. Given that [tex]\(\theta\)[/tex] is in quadrant IV where sine is negative:
[tex]\[ \sin(\theta) = -\frac{\sqrt{17}}{9} \][/tex]
4. Calculate [tex]\(\tan(\theta)\)[/tex]:
[tex]\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \][/tex]
Substitute [tex]\(\sin(\theta) = -\frac{\sqrt{17}}{9}\)[/tex] and [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex]:
[tex]\[ \tan(\theta) = \frac{-\frac{\sqrt{17}}{9}}{\frac{8}{9}} = \frac{-\sqrt{17}}{9} \times \frac{9}{8} = -\frac{\sqrt{17}}{8} \][/tex]
Therefore, the value of [tex]\(\tan(\theta)\)[/tex] is:
[tex]\[ \boxed{-\frac{\sqrt{17}}{8}} \][/tex]
1. Identify the cosine value:
[tex]\[ \cos(\theta) = \frac{8}{9} \][/tex]
2. Determine the sine value using the Pythagorean identity:
The Pythagorean identity states:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Substitute [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex] into the identity:
[tex]\[ \sin^2(\theta) + \left(\frac{8}{9}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(\theta) + \frac{64}{81} = 1 \][/tex]
[tex]\[ \sin^2(\theta) = 1 - \frac{64}{81} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{81}{81} - \frac{64}{81} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{17}{81} \][/tex]
3. Solve for [tex]\(\sin(\theta)\)[/tex]:
Since [tex]\(\sin(\theta)\)[/tex] can be positive or negative:
[tex]\[ \sin(\theta) = \pm \sqrt{\frac{17}{81}} = \pm \frac{\sqrt{17}}{9} \][/tex]
We must consider the quadrant of [tex]\(\theta\)[/tex]. Given that [tex]\(\theta\)[/tex] is in quadrant IV where sine is negative:
[tex]\[ \sin(\theta) = -\frac{\sqrt{17}}{9} \][/tex]
4. Calculate [tex]\(\tan(\theta)\)[/tex]:
[tex]\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \][/tex]
Substitute [tex]\(\sin(\theta) = -\frac{\sqrt{17}}{9}\)[/tex] and [tex]\(\cos(\theta) = \frac{8}{9}\)[/tex]:
[tex]\[ \tan(\theta) = \frac{-\frac{\sqrt{17}}{9}}{\frac{8}{9}} = \frac{-\sqrt{17}}{9} \times \frac{9}{8} = -\frac{\sqrt{17}}{8} \][/tex]
Therefore, the value of [tex]\(\tan(\theta)\)[/tex] is:
[tex]\[ \boxed{-\frac{\sqrt{17}}{8}} \][/tex]
