In any `DeltaABC`, prove that `(a-b)^(2)cos^(2)""C/2+(a+b)^(2)sin^(2)""C/2=c^(2).`

Explanation

By the projection rule in `DeltaABC`
`a=c cos B+b cos C`
`b=a cos C+cos A`
`thereforea+b=ccos B +b cos A `
`(a+b)(1-cos C)=c (cos B +cosA)`
`(a+b) (2 sin ^(2)((C)/(2)))=c (cos B+cosA)`
`(a+b)*sin^(2)((C)/(2))=c/2(cos B +cos A ) (a+b)" "...(1)`
and `a-b=c cos B+b cos C-a cos C-c cos A`
`(a-b)(1+cos C)=c (cos B -cos A) `
`(a-b)(2cos ^(2)((C)/(2)))=c (cos B -cos A)`
`(a-b)^(2)cos ^(2)((C)/(2))=C/2(cos B-cos A)(a-b)" "...(2)`
`(a-b)^(2)cos ^(2)((C)/(2))+(a+b)^(2)sin^(2)((C)/(2))=c/2[a cos B+a cos A+b cos B+`
`b cos A+a cos B-a cos A-b cos B +b cos A]` lt brgt `therefore (a-b)^(2)cos ^(2)((C)/(2))+(a+b)^(2)sin^(2)((c)/(2))`
`=C/2.2(a cos B+cos A)`
`therefore(a-b)cos ^(2)((C)/(2))+(a+b)^(2)sin^(2)((C)/(2))=c^(2)`
`[becausec=a cos B+b cos A`
Hence Provide.