Question

If `sintheta=sqrt403/22` and `pi/2<theta<pi`, how do you find `tan2theta`?

Answer 1

`tan2theta=(9sqrt403)/161`

Explanation:

If `theta in [pi/2,pi]`, `costheta<0`, so:
`costheta=-sqrt{1-sin^2theta}`
`costheta=-sqrt{1-403/484}=-sqrt{81/484}=-9/22`

`tantheta=sintheta/costheta=(sqrt{403}/22)/(-9/22)=-sqrt403/9`
`tan2theta=(2tantheta)/(1-tan^2theta)=(-(2sqrt403)/9)/(1-403/81)=(-(2sqrt403)/9)/(-322/81)=(9sqrt403)/161`