If ` (2sinalpha)/(1+cosalpha +sinalpha)=y` , then prove that `(1-cosalpha+sinalpha )/(1+sinalpha)` is also equal to y.

Explanation

Given that, ` (2sinalpha)/(1+cosalpha +sinalpha)=y`
Now, `(1-cosalpha+sinalpha )/(1+sinalpha)` = `((1-cosalpha+sinalpha) )/((1+sinalpha))*((1+cosalpha+ sinalpha))/((1+cosalpha+sinalpha))`
`" "=({(1+sinalpha)-cosalpha})/((1+sinalpha))*({(1+sinalpha)+cosalpha})/((1+cosalpha+sinalpha))`
`" "=((1+sinalpha)^(2)-cos^(2)alpha)/((1+sinalpha)(1+sinalpha+ cosalpha))`
`" "=((1+sin^(2)alpha+2sinalpha )-cos^(2)alpha)/((1+sinalpha )(1+sinalpha+cosalpha))`
`" "=(1+sin^(2)alpha+2sinalpha-1+sin^(2 )alpha)/((1+sinalpha)(1+sinalpha+cosalpha))`
`" "=(2sin^(2)alpha+2sinalpha)/((1+sinalpha)(1+sinalpha+cosalpha))`
`" "=(2sinalpha(1+sinalpha))/((1+sinalpha)(1+sinalpha+cosalpha))`
`" "=(2sinalpha)/(1+sinalpha+cosalpha)=y" "` Hence proved.