To find the value of [tex]\(\tan x\)[/tex] given that [tex]\(\sin x = \frac{12}{13}\)[/tex] and [tex]\(x\)[/tex] is an angle in Quadrant II, we can proceed as follows:
1. Determine [tex]\(\cos x\)[/tex]:
Since [tex]\(\sin x\)[/tex] has been given as [tex]\(\frac{12}{13}\)[/tex], we can use the Pythagorean identity [tex]\(\sin^2(x) + \cos^2(x) = 1\)[/tex] to find [tex]\(\cos x\)[/tex].
[tex]\[
\sin^2(x) + \cos^2(x) = 1
\][/tex]
Substitute [tex]\(\sin x = \frac{12}{13}\)[/tex]:
[tex]\[
\left(\frac{12}{13}\right)^2 + \cos^2(x) = 1
\][/tex]
Simplify the square:
[tex]\[
\frac{144}{169} + \cos^2(x) = 1
\][/tex]
Solve for [tex]\(\cos^2(x)\)[/tex]:
[tex]\[
\cos^2(x) = 1 - \frac{144}{169}
\][/tex]
[tex]\[
\cos^2(x) = \frac{169}{169} - \frac{144}{169}
\][/tex]
[tex]\[
\cos^2(x) = \frac{25}{169}
\][/tex]
Therefore, [tex]\(\cos x\)[/tex] is:
[tex]\[
\cos x = \pm \sqrt{\frac{25}{169}}
\][/tex]
[tex]\[
\cos x = \pm \frac{5}{13}
\][/tex]
Since [tex]\(x\)[/tex] is in Quadrant II, where cosine is negative, we have:
[tex]\[
\cos x = -\frac{5}{13}
\][/tex]
2. Determine [tex]\(\tan x\)[/tex]:
The tangent of an angle is given by the ratio of the sine to the cosine:
[tex]\[
\tan x = \frac{\sin x}{\cos x}
\][/tex]
Substitute the values we have:
[tex]\[
\tan x = \frac{\frac{12}{13}}{-\frac{5}{13}}
\][/tex]
Simplify the fraction:
[tex]\[
\tan x = \frac{12}{13} \times \frac{13}{-5}
\][/tex]
[tex]\[
\tan x = \frac{12}{-5}
\][/tex]
[tex]\[
\tan x = -\frac{12}{5}
\][/tex]
[tex]\[
\tan x = -2.4
\][/tex]
Thus, the value of [tex]\(\tan x\)[/tex] is [tex]\(-2.4\)[/tex].
