Certainly! Let's solve the problem step-by-step:
1. We start with the given information:
[tex]\[
\sin x = \frac{1}{\sqrt{2}}
\][/tex]
2. We will use the Pythagorean identity for trigonometric functions:
[tex]\[
\sin^2 x + \cos^2 x = 1
\][/tex]
3. Substitute [tex]\(\sin x\)[/tex] into the identity:
[tex]\[
\left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 x = 1
\][/tex]
4. Simplify the expression:
[tex]\[
\frac{1}{2} + \cos^2 x = 1
\][/tex]
5. Subtract [tex]\(\frac{1}{2}\)[/tex] from both sides to solve for [tex]\(\cos^2 x\)[/tex]:
[tex]\[
\cos^2 x = 1 - \frac{1}{2} = \frac{1}{2}
\][/tex]
6. Take the square root of both sides. Since we are considering [tex]\(0^\circ \leq x \leq 90^\circ\)[/tex], [tex]\(\cos x\)[/tex] is positive:
[tex]\[
\cos x = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
\][/tex]
7. Next, we need to find [tex]\(\tan x\)[/tex]. Recall that:
[tex]\[
\tan x = \frac{\sin x}{\cos x}
\][/tex]
8. Substitute the values of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[
\tan x = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1
\][/tex]
9. Finally, we need to find [tex]\(\cos x + \tan x\)[/tex]:
[tex]\[
\cos x + \tan x = \frac{1}{\sqrt{2}} + 1
\][/tex]
Thus, combining the results, we get:
[tex]\[
\sin x = \frac{1}{\sqrt{2}}, \quad \cos x = \frac{1}{\sqrt{2}}, \quad \tan x = 1, \quad \cos x + \tan x = 1.7071067811865475
\][/tex]
So the final answer for [tex]\(\cos x + \tan x\)[/tex] is approximately:
[tex]\[
1.707
\][/tex]
