Question 1067306
secant is negative in QII and QIII
sine is positive in QI and QII

So θ is in QII

Since {{{SECANT=HYPOTENUSE/(ADJACENT)}}} and -2 = {{{("" + 2)/(-1)}}}
we draw a right triangle in QII which has hypotenuse = +2 and
adjacent side -1

{{{drawing(100,75,-2,2,-1,2,
red(arc(0,0,1,-1,0,120)),
line(-1,0,-1,sqrt(3)),locate(-1.2,0,-1),locate(-.5,1.4,2),
line(7,0,-7,0),line(0,7,0,-7),line(0,0,-1,sqrt(3)) )}}}

By the Pythagorean theorem the opposite side is {{{sqrt(3)}}}.
It's positive because it goes up from the x-axis

{{{drawing(100,75,-2,2,-1,2,locate(-1.6,1.1,sqrt(3)),
red(arc(0,0,1,-1,0,120)),
line(-1,0,-1,sqrt(3)),locate(-1.2,0,-1),locate(-.5,1.4,2),
line(7,0,-7,0),line(0,7,0,-7),line(0,0,-1,sqrt(3)) )}}}

{{{matrix(6,5,

sin(theta),""="",OPPOSITE/HYPOTENUSE,""="",sqrt(3)/2,
cos(theta),""="",ADJACENT/HYPOTENUSE,""="",-1/2,
tan(theta),""="",OPPOSITE/ADJACENT,""="",sqrt(3)/(-1)=-sqrt(3),
csc(theta),""="",HYPOTENUSE/OPPOSITE,""="",2/sqrt(3)=2sqrt(3)/3,
sec(theta),""="",HYPOTENUSE/ADJACENT,""="",2/(-1)=-2,
cot(theta),""="",ADJACENT/OPPOSITE,""="",(-1)/sqrt(3)=-sqrt(3)/3
)}}}

Edwin</pre>