#1-cos(theta)=sin^2(theta)/(1+cos(theta))#
Recall that #sin^2(theta)+cos^2(theta)=1#. Solving for #sin^2(theta)# yields:
#sin^2(theta)=1-cos^2(theta)#
Rewrite the right side using the above identity:
#1-cos(theta)=(1-cos^2(theta))/(1+cos(theta))#
Recall the difference of squares, which states that #(x^2-a^2)=(x+a)(x-a)#
Similarly, #(a^2-x^2)=(a+x)(a-x)#
For #(1-cos^2(theta)),# we can use difference of squares where #a^2=1, x^2=cos^2(theta), a=sqrt(1)=1, x=sqrt(cos^2(theta))=cos(theta):#
#(1-cos^2(theta))=(1+cos(theta))(1-cos(theta))#
Rewrite the right side again after applying this. #(1+cos(theta))# cancels:
#1-cos(theta)=(cancel(1+cos(theta))(1-cos(theta)))/cancel(1+cos(theta))#
#1-cos(theta)=1-cos(theta)#
