Given that $\sin A = -\frac{3}{5}$ and $\cos B = -\frac{8}{17}$, where angle $A$ and angle $B$ are in the same quadrant, find the exact values of

1. Determine the quadrant for angles A and B

Since $\sin A = -\frac{3}{5}$ (negative) and $\cos B = -\frac{8}{17}$ (negative), and they are in the same quadrant, both $A$ and $B$ must be in the 3rd quadrant.

2. Find the exact values of tan A and tan B

In the 3rd quadrant, tangent is positive.
For angle $A$: opposite = 3, hypotenuse = 5 $\implies$ adjacent = $\sqrt{5^2 - 3^2} = 4$.
$\tan A = \frac{3}{4}$

For angle $B$: adjacent = 8, hypotenuse = 17 $\implies$ opposite = $\sqrt{17^2 - 8^2} = 15$.
$\tan B = \frac{15}{8}$

3. Apply the compound angle formula for tan(A + B)

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

4. Substitute the values and evaluate

$\tan(A + B) = \frac{\frac{3}{4} + \frac{15}{8}}{1 - \left(\frac{3}{4}\right)\left(\frac{15}{8}\right)}$
$= \frac{\frac{6}{8} + \frac{15}{8}}{1 - \frac{45}{32}}$
$= \frac{\frac{21}{8}}{\frac{32 - 45}{32}}$
$= \frac{\frac{21}{8}}{-\frac{13}{32}}$
$= \frac{21}{8} \times \left(-\frac{32}{13}\right)$
$= -\frac{84}{13}$

1. Determine the quadrant of angle $B$ and $\frac{B}{2}$

Since $\sin A = -\frac{3}{5} < 0$, $A$ is in the 3rd or 4th quadrant.
Since $\cos B = -\frac{8}{17} < 0$, $B$ is in the 2nd or 3rd quadrant.
Given that $A$ and $B$ are in the same quadrant, they must both be in the 3rd quadrant.
Therefore, $180^\circ < B < 270^\circ$.
Dividing by 2 gives $90^\circ < \frac{B}{2} < 135^\circ$, which means $\frac{B}{2}$ is in the 2nd quadrant.
In the 2nd quadrant, $\cos \frac{B}{2}$ is negative.

2. Apply the double angle formula for cosine to find $\cos^2 \frac{B}{2}$

Using the identity $\cos B = 2\cos^2 \frac{B}{2} - 1$:
$-\frac{8}{17} = 2\cos^2 \frac{B}{2} - 1$
$2\cos^2 \frac{B}{2} = 1 - \frac{8}{17} = \frac{9}{17}$
$\cos^2 \frac{B}{2} = \frac{9}{34}$

3. Solve for $\cos \frac{B}{2}$ and select the correct sign

$\cos \frac{B}{2} = \pm \sqrt{\frac{9}{34}} = \pm \frac{3}{\sqrt{34}}$
Since $\cos \frac{B}{2}$ must be negative:
$\cos \frac{B}{2} = -\frac{3}{\sqrt{34}} = -\frac{3\sqrt{34}}{34}$