Use the values of - 3/2 and a > 0 to find the other 5 trigonometric functions

Answer 1

sec U = -3/2

cos U = -2/3

cosine is negative in quadrants 2 and 3.

In quadrant 2:

adjacent = -2 and hypotenuse = 3

By pythagorean, the missing side is sqrt( 3^2 - (-2)^2) = sqrt(9 - 4) = sqrt(5)

sine > 0 in quadrant 2, so sin U = sqrt(5)/3

tangent <0 in quadrant 2, so tan U = sin U/ cos U = sqrt(5)/3 divided by (-2/3) = -sqrt(5)/2

secant U = 3/2, cosecant U = 3 * sqrt(5)/5 and cotangent U = -2 * sqrt(5)/5

In quadrant 3:

adjacent = -2 and hypotenuse = 3

opposite = -sqrt(5)

sin <0 ---> sin U = -sqrt(5)/3

cosine is given as -2/3 , as is the secant of -3/2

tangent>0, so tan U = sin U/cos U = -sqrt(5)/3 divided by (-2/3) = sqrt(5)/2

cosecant = -3 * sqrt(5)/5

cotangent = 2 * sqrt(5)/5

the given fact that tan a >0 tells us nothing because it is a different angle than U