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!