Answer:
A
Step-by-step explanation:
Using the trig identity
sin²x + cos²x = 1 , then cos x = ± [tex]\sqrt{1-(\frac{3\sqrt{10} }{10})^2 }[/tex]
Given
sinθ = [tex]\frac{3\sqrt{10} }{10}[/tex] , then
cosθ = ± [tex]\sqrt{1-(\frac{3\sqrt{10} }{10})^2 }[/tex]
= ± [tex]\sqrt{1-\frac{9}{10} }[/tex]
= ± [tex]\sqrt{\frac{1}{10} }[/tex]
Since θ is in second quadrant where cosθ < 0 , then
cosθ = - [tex]\frac{1}{\sqrt{10} }[/tex]
Then
tanθ = [tex]\frac{sin0}{cos0}[/tex] = [tex]\frac{\frac{3\sqrt{10} }{10} }{\frac{-1}{\sqrt{10} } }[/tex] = - 3 → A