Question

Question #47acb

Answer 1

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`