Sure, let's solve this step by step.
We are given that [tex]\(\sec \theta = \sqrt{2}\)[/tex].
1. Find [tex]\(\cos \theta\)[/tex]:
Since [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex],
[tex]\[
\cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
\][/tex]
2. Find [tex]\(\sin \theta\)[/tex]:
Using the Pythagorean identity for sine and cosine: [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex],
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{\sqrt{2}}{2}\right)^2 = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}
\][/tex]
[tex]\[
\sin \theta = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}
\][/tex]
3. Find [tex]\(\tan \theta\)[/tex]:
Recall that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex],
[tex]\[
\tan \theta = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1
\][/tex]
4. Calculate [tex]\(\frac{1 + \tan \theta}{\sin \theta}\)[/tex]:
[tex]\[
1 + \tan \theta = 1 + 1 = 2
\][/tex]
[tex]\[
\frac{1 + \tan \theta}{\sin \theta} = \frac{2}{\frac{\sqrt{2}}{2}} = 2 \times \frac{2}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2\sqrt{2}
\][/tex]
Thus, the value of [tex]\(\frac{1 + \tan \theta}{\sin \theta}\)[/tex] is [tex]\(\boxed{2\sqrt{2}}\)[/tex]. So the correct choice is (A) [tex]\(2\sqrt{2}\)[/tex].
