Here,
#cos^-1(1/x)=sin^-1((sqrt(x^2-1))/x)#
Take,
#cos^-1(1/x)=theta=>1/x=costheta=>x=1/costheta=sectheta#
Given that,
#x>=1#
#=>x>0 andx>=1#
#=>1/x>0 and 1/x<=1#
i.e. #0 < 1/x <=1=>0 < costheta <=1=>0<= theta < pi/2#
Now,
#RHS=sin^-1((sqrt(x^2-1))/x),....#where, #x=sectheta#
#=sin^-1(sqrt(sec^2theta-1)/(sectheta))#
#=sin^-1(sqrt(tan^2theta)/sectheta)....#where, #sec^2theta-1=tan^2theta#
#=sin^-1(tantheta/sectheta)#
#=sin^-1((sintheta/cancel(costheta))/(1/cancelcostheta))#
#=sin^-1(sintheta).....#where, #0<= theta < pi/2#
#=theta#
#=cos^-1(1/x)#
#=LHS#