Let's write #sec(theta) = -4.5# as #sec(theta) = -9/2#

Then:

#cos(theta) = 1/sec(theta)#

#cos(theta) = -2/9#

And:

#sin^2(theta) = 1- (-2/9)^2" [1]"#

When we take square root of both sides of equation [1], we must use #+-# on the right:

#sin(theta) = +-sqrt(1-(2/9)^2)#

We are given that #theta# is the second quadrant and we know that the sine function is positive in the second quadrant so we drop the #+-#, thereby, indicating only the positive value:

#sin(theta) = sqrt(1-(2/9)^2)#

#sin(theta) = sqrt(81/81-4/81)#

#sin(theta) = sqrt(77/81)#

#sin(theta) = sqrt(77)/9#

#tan(theta) = sin(theta)/cos(theta)#

#tan(theta) = (sqrt(77)/9)/(-2/9)#

#tan(theta) = -sqrt(77)/2#

#cot(theta) = 1/tan(theta)#

#cot(theta) = -2/sqrt(77) = -2sqrt77/77#

#csc(theta) = 1/sin(theta)#

#csc(theta) = 9/sqrt77#

#csc(theta) = 9sqrt77/77#