#sin theta = 8/17 ; pi/2< theta< pi :. theta # is in 2nd quadrant
#Sin theta = P/H ; P=8 , H=17 :. B^2= H^2-P^2= 17^2-8^2#
#:. B^2=225 :. B=15 , Cos theta = -B/H= -15/17 ; cos theta#
is negative in 2nd quadrant.
# sin theta=2sin (theta/2) cos (theta/2)=8/17#
# cos theta=cos (2*theta/2)=2cos^2(theta/ 2)-1=-15/17 #
#2cos^2(theta/ 2)=-15/17+1=2/17:. #
#cos^2(theta/ 2)=1/17:. cos(theta/ 2)=1/sqrt17 :.#
# 2sin (theta/2) cos (theta/2)=8/17# or
# 2sin (theta/2)*1/sqrt17=8/17# or
# 2sin (theta/2)=8/17*sqrt17# or
# sin (theta/2)=4/17*sqrt17= 4/sqrt17#
#:. csc (theta/2)=1/sin (theta/2)=sqrt17/4# [Ans]