Question 754199
{{{sec(theta)=4}}}, 



{{{cos(theta)=1/4}}}





{{{3pi/2<theta<2pi}}} 


i.e. we are in quadrant IV


means...


{{{sin(theta)=-sqrt(15)/4}}}

{{{sin2*theta= 2sin(theta)cos(theta)=2(-sqrt(15)/4)(1/4)=-sqrt(15)/8}}}



{{{cos2*theta=2cos^2(theta)-1=2(1/4)-1=-1/2}}} 



We forgo using tangent identities since we don't need them

{{{tan2*theta= sin(2theta)/cos(2theta)=(-sqrt(15)/8)/(-1/2)=-sqrt(15)/4}}}