If tan(θ) is negative, then θ must be either in Q-II or else in Q-IV.
Fortunately, the question tells us that it's in Q-II.
If you draw a circle on the x- and y-axes, then draw a right triangle
in Q-II, then mark the legs 3 and -2, then the hypotenuse of the
triangle ... also the radius of the circle ... is √13 .
Look for the angle whose tangent is -3/2.
tangent = (opposite) / (adjacent)
So the side opposite is the 3 and the side adjacent is the -2.
For that same angle, cosine = (adjacent) / (hypotenuse) .
The adjacent side is still the -2, and the hypotenuse is √13 .
So the cosine of the same angle is
- 2 / √13 .
To rationalize the denominator (get that square root out of there),
multiply top and bottom by √13 . Then you have
(- 2 / √13) · (√13 / √13)
= - 2 √13 / 13 .
