Question

Given `sinx=sqrt3/2, cosx=-1/2`, how do you find the remaining trigonometric functions?

Answer 1

`tanx=-sqrt3`

`secx=-2`

`cscx=2/3sqrt3`

`cotx=-sqrt3/3`

Explanation:

Since `tantheta=sintheta/costheta`, you get

`tanx=(sqrt3/cancel2)/(-1/cancel2)=-sqrt3`

Since `sectheta=1/costheta`, you get

`secx=1/(-1/2)=-2`

Since `csctheta=1/sintheta`, you get

`cscx=1/(sqrt3/2)=2/sqrt3=2/3sqrt3`

Since `cottheta=1/tantheta`, you get

`cotx=1/-sqrt3=-sqrt3/3`