ANSWER
[tex]\tan(\pi + \theta)= \frac{5}{12} [/tex]
EXPLANATION
We first obtain
[tex] \cos( \theta) [/tex]
using the Pythagorean Identity.
[tex]\cos ^{2} ( \theta) + \sin ^{2} ( \theta) = 1[/tex]
[tex] \implies \: \cos ^{2} ( \theta) + ( - \frac{5}{13} )^{2} = 1[/tex]
[tex]\implies \: \cos ^{2} ( \theta) + \frac{25}{169}= 1[/tex]
[tex]\implies \: \cos ^{2} ( \theta) = 1 - \frac{25}{169}[/tex]
[tex]\implies \: \cos ^{2} ( \theta) = \frac{144}{169}[/tex]
[tex]\implies \: \cos ( \theta) = \pm \: \sqrt{\frac{144}{169} } [/tex]
[tex]\implies \: \cos ( \theta) = \pm \: \frac{12}{13} [/tex]
In the third quadrant, the cosine ratio is negative.
[tex]\implies \: \cos ( \theta) = - \: \frac{12}{13} [/tex]
The tangent function has a period of π and [tex]\pi + \theta[/tex] is in the third quadrant.
This implies that:
[tex] \tan(\pi + \theta)= \tan( \theta) [/tex]
[tex]\tan(\pi + \theta)= \frac{ \sin( \theta) }{ \cos( \theta) } [/tex]
[tex]\tan(\pi + \theta)= \frac{ - \frac{ 5}{13} }{ - \frac{12}{13} } [/tex]
This gives us:
[tex]\tan(\pi + \theta)= \frac{5}{12} [/tex]
