Find [tex]$\sec \theta$[/tex] if [tex]$\theta$[/tex] is in Quadrant IV and [tex]$\sin \theta=-\frac{1}{5}$[/tex].

A. [tex][tex]$\frac{2 \sqrt{6}}{5}$[/tex][/tex]
B. [tex]$\frac{5 \sqrt{6}}{12}$[/tex]
C. [tex]$\frac{2 \sqrt{6}}{5}$[/tex]
D. [tex][tex]$-\frac{5 \sqrt{6}}{12}$[/tex][/tex]



Answer :

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]