Answer:
[tex]\sec \theta = \frac{\sqrt{22}}{4}[/tex]
Step-by-step explanation:
Given
[tex]\tan^2 \theta = \frac{3}{8}[/tex]
Required
[tex]\sec\ \theta[/tex]
We have:
[tex]\sec^2\theta = 1 + \tan^2 \theta[/tex]
This gives:
[tex]\sec^2\theta = 1 + \frac{3}{8}[/tex]
Take lcm and solve
[tex]\sec^2\theta = \frac{9+3}{8}[/tex]
[tex]\sec^2\theta = \frac{11}{8}[/tex]
Take square roots
[tex]\sec \theta = \frac{\sqrt{11}}{\sqrt 8}[/tex]
[tex]\sec \theta = \frac{\sqrt{11}}{2\sqrt 2}[/tex]
Rationalize
[tex]\sec \theta = \frac{\sqrt{11}}{2\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}[/tex]
[tex]\sec \theta = \frac{\sqrt{22}}{4}[/tex]
