How do you prove: #sin^2x/(1-cosx) = 1+cosx#?

1 Answer
Jun 12, 2016

Modifying just the left-hand side:

We can use the Pythagorean Identity to rewrite #sin^2x#. The Pythagorean Identity states that

#color(red)(barul|color(white)(a/a)color(black)(sin^2x+cos^2x=1)color(white)(a/a)|)#

Which can be rearranged to say that

#color(teal)(barul|color(white)(a/a)color(black)(sin^2x=1-cos^2x)color(white)(a/a)|)#

Thus, we see that

#sin^2x/(1-cosx)=(1-cos^2x)/(1-cosx)#

We can now factor the numerator of this fraction. #1-cos^2x# is a difference of squares, which can be factored as:

#color(blue)(barul|color(white)(a/a)color(black)(a^2-b^2=(a+b)(a-b))color(white)(a/a)|)#

This can be applied to #1-cos^2x# as follows:

#color(orange)(barul|color(white)(a/a)color(black)(1-cos^2x=1^2-(cosx)^2=(1+cosx)(1-cosx))color(white)(a/a)|)#

Therefore,

#(1-cos^2x)/(1-cosx)=((1+cosx)(1-cosx))/(1-cosx)#

Since there is #1-cosx# present in both the numerator and denominator, it can be cancelled:

#((1+cosx)(1-cosx))/(1-cosx)=((1+cosx)color(red)(cancel(color(black)((1-cosx)))))/(color(red)(cancel(color(black)((1-cosx)))))=1+cosx#

This is what we initially set out to prove.