Let us use the Identity : tantheta=(2tan(theta/2))/(1-tan^2(theta/2)
Letting, tan(theta/2)=t, tantheta=(2t)/(1-t^2).............(1).
Since, tantheta=2, we have, by (1),
2=(2t)/(1-t^2)rArr1-t^2=t, or, t^2+t-1+0
t=(-1+-sqrt(1^2-4*1*(-1)))/2=(-1+-sqrt5)/2
Let us note that, by data, 0<,theta<,pi/2rArr 0<,theta/2<,pi/4
:. theta/2 lies in the First Quadrant, in which, tan(theta/2)>0,
and, also, sin(theta/2)>0
This means that tan(theta/2)!=(-1-sqrt5)/2
:. tan(theta/2)=(sqrt5-1)/2rArr cot(theta/2)=2/(sqrt5-1)
Now, csc^2(theta/2)=1+cot^2(theta/2)=1+4/(sqrt5-1)^2
:.csc^2(theta/2)={(5-2sqrt5+1)+4}/(sqrt5-1)^2=(10-2sqrt5)/(sqrt5-1)^2
:.sin(theta/2)=+(sqrt5-1)/sqrt(10-2sqrt5)={(sqrt5-1)sqrt(10+2sqrt5)}/{sqrt(10-2sqrt5)sqrt(10+2sqrt5)}
={(sqrt5-1)sqrt(10+2sqrt5)}/sqrt(100-20)={(sqrt5-1)sqrt(10+2sqrt5)}/(4sqrt5)
={(sqrt5-1)(sqrt5)sqrt(10+2sqrt5)}/(4*5)={(sqrt5-1)sqrt(50+10sqrt5)}/20