Question

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

Answer 1

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.