Answer:
tan θ = -√7/3
Step-by-step explanation:
We have the equation as following:
[tex]sin^{2} x + cos^{2} x =1[/tex]
=> (cos x)^ 2 = 1 - (sin x)^2
So that:
(cos θ)^2 = 1 - (sin θ)^2 = 1- (√7/4)^2 = 1 - 7/16 = 9/16
=> (cos θ)^2 = 9/16
We have the equation:
[tex]tan^{2} x = \frac{1}{cos^{2}x } -1[/tex]
=> (tan θ)^2 = 1/(cos θ)^ 2 - 1 = 1/(9/16 ) - 1 = 16/9 - 1 = 7/9
=> tan θ = √7/3 or tan θ = -√7/3
As θ is in the quadrant II, so that tan θ would have the negative value
=> tan θ = -√7/3
Answer C is correct
