To solve the problem, we need to find \(\tan(A - B)\) given \(\sin(A) = \frac{5}{13}\) with \(\frac{\pi}{2} < A < \pi\) and \(\tan(B) = -\sqrt{13}\) with \(\frac{\pi}{2} < B < \pi\).
Here's the detailed solution step-by-step:
1. Determine \(\cos(A)\):
Given \(\sin(A) = \frac{5}{13}\), we can use the Pythagorean identity \(\sin^2(A) + \cos^2(A) = 1\) to find \(\cos(A)\):
[tex]\[
\cos^2(A) = 1 - \sin^2(A) = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}
\][/tex]
Since \(\frac{\pi}{2} < A < \pi\), \(A\) is in the second quadrant where \(\cos(A)\) is negative:
[tex]\[
\cos(A) = -\sqrt{\frac{144}{169}} = -\frac{12}{13}
\][/tex]
2. Calculate \(\tan(A)\):
Given \(\tan(A) = \frac{\sin(A)}{\cos(A)}\):
[tex]\[
\tan(A) = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12}
\][/tex]
3. Use the tangent subtraction formula:
The formula for \(\tan(A - B)\) is:
[tex]\[
\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A) \tan(B)}
\][/tex]
Substituting \(\tan(A) = -\frac{5}{12}\) and \(\tan(B) = -\sqrt{13}\):
[tex]\[
\tan(A - B) = \frac{-\frac{5}{12} - (-\sqrt{13})}{1 + \left(-\frac{5}{12}\right) \left(-\sqrt{13}\right)}
\][/tex]
4. Simplify the expression:
[tex]\[
\tan(A - B) = \frac{-\frac{5}{12} + \sqrt{13}}{1 + \frac{5 \sqrt{13}}{12}}
\][/tex]
5. Combine terms in both numerator and denominator:
[tex]\[
\tan(A - B) = \frac{\sqrt{13} - \frac{5}{12}}{1 + \frac{5 \sqrt{13}}{12}} = \frac{\sqrt{13} - \frac{5}{12}}{\frac{12 + 5 \sqrt{13}}{12}} = \frac{12(\sqrt{13} - \frac{5}{12})}{12 + 5 \sqrt{13}}
\][/tex]
[tex]\[
\tan(A - B) = \frac{12\sqrt{13} - 5}{12 + 5 \sqrt{13}}
\][/tex]
From the options provided, the equivalent fraction form is:
[tex]\[
\boxed{\frac{12 + 5 \sqrt{13}}{-5 + 12 \sqrt{13}}}
\][/tex]
However, taking the negative denominator into account simplifies to:
[tex]\[
\tan(A - B) = 1.274374775947988
\][/tex]
Thus, [tex]\(\tan(A - B) = \boxed{\frac{-5 + 12 \sqrt{13}}{12 + 5 \sqrt{13}}}\)[/tex] is not necessarily in your provided list but matches the computed result.
