Prove

`2cos^2(theta/2) = (sin^2theta)/(1-costheta)`

The half angle identity states:

`cos(theta/2) = sqrt((1+costheta)/2)`

This turns out left side into,

`2(1+costheta)/2 =(sin^2theta)/(1-costheta)`

`1+costheta =(sin^2theta)/(1-costheta)`

`(1-costheta)/(1-costheta) (1+costheta) =(sin^2theta)/(1-costheta)`

This simplifies into

`(1-cos^2theta)/(1-costheta) =(sin^2theta)/(1-costheta)`

Which turns into

`(sin^2theta)/(1-costheta) =(sin^2theta)/(1-costheta)`

It is proven!

Using the following identities:

`sin^2 x + cos^2 x =1`

and `cos^2 x = (1+cos 2x)/2`

Left hand side = 2 cos^2 (theta/2) = 2. (1+cos 2.theta/2)/2 = 1+cos theta

multiplying and dividing by (1-cos theta), we get.

LHS = (1+cos theta). (1-cos theta)/ (1-cos theta) = (1-cos^2 theta)/(1-cos theta)

= (sin^2 theta + cos^2 theta - cos theta)/(1cos theta) = sin^2 theta/ (1-cos theta). = RHS

Hence proved.

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