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If [tex]\sin x = \frac{12}{13}[/tex], where [tex]x[/tex] is a Quadrant II angle, find the value of [tex]\tan x[/tex].

Sagot :

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].