To solve for [tex]\(\tan(\theta)\)[/tex] given that [tex]\(\csc(\theta) = -\sqrt{5}\)[/tex] and [tex]\(\theta\)[/tex] is in Quadrant IV, we will follow a systematic approach with the trigonometric identities and the properties of trigonometric functions in different quadrants.
1. Identify [tex]\(\sin(\theta)\)[/tex]:
We know that:
[tex]\[
\csc(\theta) = \frac{1}{\sin(\theta)}
\][/tex]
Given that [tex]\(\csc(\theta) = -\sqrt{5}\)[/tex], we can find [tex]\(\sin(\theta)\)[/tex] as:
[tex]\[
\sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{-\sqrt{5}} = -\frac{1}{\sqrt{5}}
\][/tex]
Rationalizing the denominator, we get:
[tex]\[
\sin(\theta) = -\frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}
\][/tex]
2. Determine [tex]\(\cos(\theta)\)[/tex]:
We use the Pythagorean identity:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
Substituting [tex]\(\sin(\theta)\)[/tex] into the equation, we get:
[tex]\[
\left(-\frac{\sqrt{5}}{5}\right)^2 + \cos^2(\theta) = 1
\][/tex]
[tex]\[
\frac{5}{25} + \cos^2(\theta) = 1
\][/tex]
[tex]\[
\frac{1}{5} + \cos^2(\theta) = 1
\][/tex]
Subtract [tex]\(\frac{1}{5}\)[/tex] from both sides:
[tex]\[
\cos^2(\theta) = 1 - \frac{1}{5} = \frac{4}{5}
\][/tex]
Taking the positive square root (since [tex]\(\theta\)[/tex] is in Quadrant IV and [tex]\(\cos\)[/tex] is positive in Quadrant IV):
[tex]\[
\cos(\theta) = \sqrt{\frac{4}{5}} = \frac{2\sqrt{5}}{5}
\][/tex]
3. Calculate [tex]\(\tan(\theta)\)[/tex]:
We know the definition of [tex]\(\tan(\theta)\)[/tex]:
[tex]\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\][/tex]
Substituting the values of [tex]\(\sin(\theta)\)[/tex] and [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\tan(\theta) = \frac{-\frac{\sqrt{5}}{5}}{\frac{2\sqrt{5}}{5}} = \frac{-\sqrt{5}}{5} \cdot \frac{5}{2\sqrt{5}} = -\frac{1}{2}
\][/tex]
Thus, the exact form of [tex]\(\tan(\theta)\)[/tex] is:
[tex]\[
\tan(\theta) = -\frac{1}{2}
\][/tex]
