ANSWER
[tex] \tan(x + y) = - \frac{63}{16} [/tex]
EXPLANATION
We were given that,
[tex] \csc(x) = \frac{5}{3} [/tex]
This implies that,
[tex] \sin(x) = \frac{3}{5} [/tex]
We use the Pythagorean identity
[tex] \sin^{2} (x) + \cos^{2} (x)= 1[/tex]
to get,
[tex] \cos(x) = \sqrt{1 - ( { \frac{3}{5} })^{2}} = \frac{4}{5} [/tex]
We were also given that,
[tex] \cos(y) = \frac{5}{13} [/tex]
This means that,
[tex] \sin(y) = \sqrt{1 - {( \frac{5}{13}) }^{2} } = \frac{12}{13} [/tex]
This is because,
[tex]0 < \: x \: < \frac{\pi}{2} [/tex]
[tex]0 < \: y \: < \frac{\pi}{2} [/tex]
This angles are in the first quadrant so we pick the positive values.
[tex] \tan(x + y) = \frac{ \sin(x + y) }{ \cos(x + y) } [/tex]
[tex] \tan(x + y) = \frac{ \sin(x ) \cos(y) + \sin(y) \cos(x) }{ \cos(x) \cos(y) - \sin(x) \sin(y) } [/tex]
[tex] \tan(x + y) = \frac{ \frac{3}{5} \times \frac{5}{13} + \frac{12}{13} \times \frac{4}{5} }{ \frac{4}{5} \times \frac{5}{13} - \frac{3}{5} \times \frac{12}{13} } [/tex]
[tex] \tan(x + y) = - \frac{63}{16} [/tex]
The correct answer is D
