To find [tex]\(\sec \theta\)[/tex] given that [tex]\(\theta\)[/tex] is in Quadrant IV and [tex]\(\sin \theta = -\frac{1}{5}\)[/tex], we can follow these steps:
1. Identify the trigonometric identity involving sine and cosine:
We use the Pythagorean identity:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
2. Substitute the given value of [tex]\(\sin \theta\)[/tex] into the identity:
Given [tex]\(\sin \theta = -\frac{1}{5}\)[/tex],
[tex]\[
\left( -\frac{1}{5} \right)^2 + \cos^2 \theta = 1
\][/tex]
Simplify the squared term:
[tex]\[
\frac{1}{25} + \cos^2 \theta = 1
\][/tex]
3. Solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[
\cos^2 \theta = 1 - \frac{1}{25}
\][/tex]
[tex]\[
\cos^2 \theta = \frac{25}{25} - \frac{1}{25}
\][/tex]
[tex]\[
\cos^2 \theta = \frac{24}{25}
\][/tex]
4. Determine [tex]\(\cos \theta\)[/tex] by taking the square root:
Since [tex]\(\theta\)[/tex] is in Quadrant IV, cosine is positive:
[tex]\[
\cos \theta = \sqrt{\frac{24}{25}}
\][/tex]
[tex]\[
\cos \theta = \frac{\sqrt{24}}{5}
\][/tex]
Simplify [tex]\(\sqrt{24}\)[/tex]:
[tex]\[
\cos \theta = \frac{2\sqrt{6}}{5}
\][/tex]
5. Find [tex]\(\sec \theta\)[/tex]:
Recall that [tex]\(\sec \theta\)[/tex] is the reciprocal of [tex]\(\cos \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\cos \theta}
\][/tex]
Substitute the value of [tex]\(\cos \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\frac{2\sqrt{6}}{5}}
\][/tex]
Simplify by multiplying by the reciprocal:
[tex]\[
\sec \theta = \frac{5}{2\sqrt{6}}
\][/tex]
To rationalize the denominator, multiply the numerator and the denominator by [tex]\(\sqrt{6}\)[/tex]:
[tex]\[
\sec \theta = \frac{5 \sqrt{6}}{2 \cdot 6}
\][/tex]
Simplify the fraction:
[tex]\[
\sec \theta = \frac{5 \sqrt{6}}{12}
\][/tex]
Hence, the required value is:
[tex]\[
\boxed{\frac{5 \sqrt{6}}{12}}
\][/tex]
