The value of [tex]\sec\theta =\pm\sqrt{\frac{11}{8}}[/tex].
Solution:
Given data:
[tex]$\tan^2\theta=\frac{3}{8}[/tex]
To find the value of [tex]\sec\theta:[/tex]
Using trigonometric identity,
[tex]\sec^2\theta=1+\tan^2\theta[/tex]
Substitute [tex]\tan^2\theta=\frac{3}{8}[/tex] in the identity, we get
[tex]$\Rightarrow \sec^2\theta=1+\frac{3}{8}[/tex]
1 can be written as [tex]\frac{1}{1}[/tex].
[tex]$ =\frac{1}{1} +\frac{3}{8}[/tex]
Do cross multiplication.
[tex]$ =\frac{8}{8} +\frac{3}{8}[/tex]
Denominators are same, so you can add the fractions.
[tex]$ =\frac{8+3}{8}[/tex]
[tex]$\Rightarrow \sec^2\theta =\frac{11}{8}[/tex]
Taking square root on both sides, we get
[tex]$\Rightarrow \sec\theta =\pm\sqrt{\frac{11}{8}}[/tex]
Option B is the correct answer.
Hence the value of [tex]\sec\theta =\pm\sqrt{\frac{11}{8}}[/tex].
