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#
