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( cos 2x - cos 2alpha)/(cos x - cos alph...

` ( cos 2x - cos 2alpha)/(cos x - cos alpha)`

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Here, we will use ,
`cosC-cosD = 2sin((C+D)/2)sin((D-C)/2)`
`(cos2x-cos2alpha)/(cosx-cosalpha)`
`=(2sin(x+alpha)sin(alpha-x))/(2sin((x+alpha)/2)sin((alpha-x)/2))`
`=(sin(x+alpha)sin(alpha-x))/(sin((x+alpha)/2)sin((alpha-x)/2))`
`=((2sin((x+alpha)/2)cos((x+alpha)/2))(2sin((alpha-x)/2)cos((alpha-x)/2)))/((sin((x+alpha)/2)sin((alpha-x)/2))`
`=4cos((x+alpha)/2)cos((x-alpha)/2)` ( As `cos(-y) = cosy`)
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