We are tasked with finding $\tan(A - B)$, where:

• $\sin(A) = \frac{5}{13}$ and $\frac{\pi}{2} < A < \pi$,
• $\tan(B) = -\sqrt{13}$ and $\frac{\pi}{2} < B < \pi$.

Let's solve this step by step:

Step 1: Determine $\cos(A)$

Since $\frac{\pi}{2} < A < \pi$, angle $A$ is in the second quadrant, where sine is positive and cosine is negative. We can use the Pythagorean identity:

$\sin^2(A) + \cos^2(A) = 1$ $\left(\frac{5}{13}\right)^2 + \cos^2(A) = 1$ $\frac{25}{169} + \cos^2(A) = 1$ $\cos^2(A) = 1 - \frac{25}{169} = \frac{144}{169}$ $\cos(A) = -\frac{12}{13}$ We take the negative root because cosine is negative in the second quadrant.

Step 2: Determine $\sin(B)$ and $\cos(B)$

For angle $B$, we know $\tan(B) = -\sqrt{13}$ and $B$ is in the second quadrant. The tangent identity is:

$\tan(B) = \frac{\sin(B)}{\cos(B)}$

Since $\tan(B) = -\sqrt{13}$, we can set up the following relation:

$\frac{\sin(B)}{\cos(B)} = -\frac{\sqrt{13}}{1}$

Using the Pythagorean identity again:

$\sin^2(B) + \cos^2(B) = 1$

Let's assume $\cos(B) = x$, then $\sin(B) = -\sqrt{13}x$ (since tangent is sine over cosine). Substituting these into the Pythagorean identity:

$(-\sqrt{13}x)^2 + x^2 = 1$ $13x^2 + x^2 = 1$ $14x^2 = 1$ $x^2 = \frac{1}{14}, \quad x = \frac{1}{\sqrt{14}}$ Thus, $\cos(B) = \frac{-1}{\sqrt{14}}$ (negative because cosine is negative in the second quadrant). Now, $\sin(B)$ becomes:

$\sin(B) = -\sqrt{13} \times \frac{-1}{\sqrt{14}} = \frac{\sqrt{13}}{\sqrt{14}}$

Step 3: Use the tangent subtraction formula

The formula for $\tan(A - B)$ is:

$\tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$

First, calculate $\tan(A)$:

$\tan(A) = \frac{\sin(A)}{\cos(A)} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12}$

Now substitute $\tan(A) = -\frac{5}{12}$ and $\tan(B) = -\sqrt{13}$:

$\tan(A - B) = \frac{-\frac{5}{12} - (-\sqrt{13})}{1 + \left(-\frac{5}{12}\right)(-\sqrt{13})}$ $\tan(A - B) = \frac{-\frac{5}{12} + \sqrt{13}}{1 + \frac{5\sqrt{13}}{12}}$ Multiply numerator and denominator by 12 to simplify:

$\tan(A - B) = \frac{-5 + 12\sqrt{13}}{12 + 5\sqrt{13}}$

Step 4: Compare with the given options

This matches the fourth option:

$\boxed{\frac{-5 + 12\sqrt{13}}{12 + 5\sqrt{13}}}$

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Would you like further clarification or have any questions?

Here are 5 related questions:

1. How can you find $\sin(A + B)$ given $\sin(A)$ and $\tan(B)$?
2. What is the range of possible values for $\tan(A - B)$?
3. Can you solve for $\cos(A - B)$ using the same information?
4. How do quadrant signs affect trigonometric functions in general?
5. How would the solution change if $A$ or $B$ were in the first quadrant?

Tip: When working with trigonometric identities, always check which quadrant your angles are in to get the correct signs for sine, cosine, and tangent.