Question

Please help me right now.

Answer 1

Look at the picture.

Definitions:

$\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}\\\\\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{r}{x}\\\\\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{r}{y}$

We have:

$\sec\theta=3=\dfrac{3}{1}$ and Quadrant I (x > 0, y > 0)

Therefore r = 3 and x = 1. Calculate y:

$r=\sqrt{x^2+y^2}\to\sqrt{1^2+y^2}=3\ \ \ |^2\\\\1+y^2=9\ \ \ |-1\\\\y^2=8\to y=\sqrt8\\\\y=\sqrt{4\cdot2}\to y=\sqrt4\cdot\sqrt2\to y=2\sqrt2$

Substitute to the formula of tan:

$\tan\theta=\dfrac{2\sqrt2}{1}=2\sqrt2$

Answer: a.