ANSWER
The correct answer is C
[tex] \sin ( \theta) = - \frac{ \sqrt{33} }{7} , \tan ( \theta) = - \frac{ \sqrt{33} }{ 4 } [/tex]
EXPLANATION
It was given that,
[tex] \cos( \theta) = \frac{4}{7} [/tex]
and
[tex] \csc( \theta) < 0[/tex]
This means that,
[tex] \theta[/tex]
is in the fourth quadrant.
We use the identity,
[tex] \cos ^{2} ( \theta) + \sin ^{2} ( \theta) = 1[/tex]
This implies that,
[tex] ( { \frac{4}{7} })^{2} + \sin ^{2} ( \theta) = 1[/tex]
[tex] { \frac{16}{49} } + \sin ^{2} ( \theta) = 1[/tex]
[tex] \sin ^{2} ( \theta) = 1 - { \frac{16}{49} }[/tex]
[tex] \sin ^{2} ( \theta) = { \frac{33}{49} }[/tex]
[tex] \sin ( \theta) = \pm \sqrt{{ \frac{33}{49} }} [/tex]
[tex] \sin ( \theta) = \pm \frac{ \sqrt{33} }{7} [/tex]
But
[tex] \csc( \theta) < 0[/tex]
This implies that,
[tex] \sin ( \theta) = - \frac{ \sqrt{33} }{7} [/tex]
[tex] \tan ( \theta) = \frac{ \sin( \theta) }{ \cos( \theta) } [/tex]
[tex] \tan ( \theta) = \frac{ - \frac{ \sqrt{33} }{7} }{ \frac{4}{7} } [/tex]
[tex] \tan ( \theta) = - \frac{ \sqrt{33} }{ 4 } [/tex]
