`sec(u) = -2, pi |

Given `sec(u)=-2, pi<u<(3pi)/2`

Angle u is in quadrant 3. If `sec(u)=-2` then `cos(u)=-1/2.`

If a right triangle is drawn in quadrant 3, the side adjacent to angle u would be 1 and the hypotenuse of the triangle would be 2. Using the pythagorean theorem the third side  would be `sqrt(3).`

`sin(2u)=2sin(u)cos(u)=2(-sqrt3/2)(-1/2)=(2sqrt3)/4=sqrt3/2`

`cos(2u)=2cos^2u-1=2(-1/2)^2-1=2(1/4)-1=-1/2`

`tan(2u)=(2tan(u))/(1-tan^2(u))=((2sqrt3)/1)/(1-(sqrt3/1)^2)=(2sqrt3)/(1-3)=(2sqrt3)/-2=-sqrt3`

Given `sec(u)=-2, pi<u<(3pi)/2`

Angle u is in quadrant 3. If `sec(u)=-2` then `cos(u)=-1/2.`

If a right triangle is drawn in quadrant 3, the side adjacent to angle u would be 1 and the hypotenuse of the triangle would be 2. Using the pythagorean theorem the third side  would be `sqrt(3).`

`sin(2u)=2sin(u)cos(u)=2(-sqrt3/2)(-1/2)=(2sqrt3)/4=sqrt3/2`

`cos(2u)=2cos^2u-1=2(-1/2)^2-1=2(1/4)-1=-1/2`

`tan(2u)=(2tan(u))/(1-tan^2(u))=((2sqrt3)/1)/(1-(sqrt3/1)^2)=(2sqrt3)/(1-3)=(2sqrt3)/-2=-sqrt3`