Question

PLEASE PLEASE HELP ME! Brainliest answer for first correct answer...... PLEASE!

Answer 1

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  .

Answer 2

-3/(sqrt)3
First you have to draw it out in the second quadrant, and then when you write in the tangent (opposite/adjacent) you can use Pythagorean theorem to solve for the last number (hypotenuse) and then cos is adjacent over hypotenuse