Look at the picture.
Definitions:
[tex] \sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}\\\\\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{r}{x}\\\\\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{r}{y} [/tex]
We have:
[tex]\sec\theta=3=\dfrac{3}{1}[/tex] and Quadrant I (x > 0, y > 0)
Therefore r = 3 and x = 1. Calculate y:
[tex]r=\sqrt{x^2+y^2}\to\sqrt{1^2+y^2}=3\ \ \ |^2\\\\1+y^2=9\ \ \ |-1\\\\y^2=8\to y=\sqrt8\\\\y=\sqrt{4\cdot2}\to y=\sqrt4\cdot\sqrt2\to y=2\sqrt2[/tex]
Substitute to the formula of tan:
[tex]\tan\theta=\dfrac{2\sqrt2}{1}=2\sqrt2[/tex]
Answer: a.
