Question

What is the exact value of sec A if it passes through P(2,-2) in standard position?

Answer 1

What is the exact value of $\sec \alpha$ if it passes through $A(2, -2)$ in standard position?

Solution:

So, we need to find

$$\sec (\angle BOA) = \frac {1}{\cos (\angle BOA)}$$

The point $A$ is situated in the $4^{\text{th}}$ quadrant, so the value of cosine is positive.

From the triangle $OBA$ :

$$OB = BA = 2$$

The triangle $OBA$ is $\Delta => \angle BOA = \angle BAO = 45{}^{\circ}$

$$\cos (\angle BOA) = \cos (45{}^{\circ}) = \frac {\sqrt {2}}{2}$$ $$\sec (\angle BOA) = \frac {1}{\cos (\angle BOA)} = \frac {1}{\frac {\sqrt {2}}{2}} = \frac {2}{\sqrt {2}} = \sqrt {2}$$

Answer: $\sqrt{2}$