Recall that, #1-cos2theta=2sin^2theta, and (1+cos2theta)=2cos^2theta#.
#:. y=ln((1-cosx)/(1+cosx))^(1/2)#,
#=ln{(2sin^2(x/2))/(2cos^2(x/2))}^(1/2)#,
#=ln(sin(x/2)/cos(x/2))............(star^star)#,
#rArr y=lnsin(x/2)-lncos(x/2)#.
#:. dy/dx=d/dx{lnsin(x/2)}-d/dx{lncos(x/2)}......(star)#.
Here, by the Chain Rule,
#d/dx{lnsin(x/2)}=1/sin(x/2)d/dx{sin(x/2)}#,
#=1/sin(x/2)*cos(x/2)d/dx{x/2}#,
#rArr d/dx{lnsin(x/2)}=1/2*cos(x/2)/sin(x/2).............(star^1)#.
Similarly, #d/dx{lncos(x/2)}=-1/2*sin(x/2)/cos(x/2)........(star^2)#.
Utilising #(star^1) and (star^2)" in "(star),# we have,
#dy/dx=1/2{sin(x/2)/cos(x/2)-(-cos(x/2)/sin(x/2))}#,
#={sin^2(x/2)+cos^2(x/2)}/{2sin(x/2)cos(x/2)}#,
# rArr dy/dx=1/sinx=cscx#.
Aliter :
By, #(star^star), y=lntan(x/2)#.
#:. dy/dx=1/tan(x/2)*d/dx{tan(x/2)},#
#=1/tan(x/2)*sec^2(x/2)*d/dx{x/2},#
#=cos(x/2)/sin(x/2)*sin^2(x/2)/cos^2(x/2)*1/2#,
#=1/(2sin(x/2)cos(x/2))#.
# rArr dy/dx=cscx,# as before!
Enjoy Maths., and Spread the Joy!
