If tan^2 A = 5/9 and A is between 3pi/2 and 2pi what is the value of cos A

So A is in quadrant 4 so tanA < 0

tanA = -√(5)/3

and tanA = opposite/adjacent

so opposite = √5
adjacent = 3
Hypotenuse = √(5 + 9) = √(14)

cosA = adjacent/hypotenuse

cosA = 3/√(14) = 3√(14)/14

First, let's solve for A (which is an angle between 3pi/2 and 2pi) [270 degrees - 360 degrees]

tan^2 A = 5/9
tan A = rt5/ 3 1) square both sides
arctan (tan A) = arctan (rt5/ 3) 2) take the arctan/inverse tangent of both sides
A = arctan (rt 5/3) 3) This is what A is, believe it or not, arctan (rt5/3) is an ANGLE!
Note: Since A is in the 4th quadrant, and cos (or x values) are positive in the 4th quadrant, we don't have to change anything.

cos (arctan (rt5/ 3)) should be the answer.

Note: The person above me got the same answer, do whichever method you feel most comfortable with.

cosA = 3/√(14) = 3√(14)/14 = cos (arctan (rt(5)/3))

sec^2(A) = 1 + tan^2(A) [Identity]
sec^2(A) = 1 + (5/9)^2
sec^2(A) = 1 + 25/81 =(81 + 25)/81 = 106/81
sec(A) = sqrt(106)/9
Now cos(A) = 1/sec(A) = 1/[sqrt(106)/9] = 9/sqrt(106)