Given that csc θ = -4 and tan θ > 0, find the exact value of

Question

a. x
b. sin θ
c. cos θ
d. r
e. sec θ
f. cot θ

Answer 1

same as the others ....

cscØ = -4
so sinØ = -1/4

we also know that tanØ>0 so
the sine is negative and the tangent is positive ---> Ø must be in III

since sinØ = -1/4, in III, y = -1, r = 4
by Pythagoras and the CAST rule
x = -√15

The rest is now easy, just use your definitions.
let me know what you get.

e.g.
cotØ = x/y = -√15/-1 = √15