LHS `=cot4x(sin5x+sin3x)`
`=(cos4x)/(sin4x)[2.sin(5x+3x)/(2) cos(5x-3x)/(2)]`
`=(cos4x)/(sin4x).2sin4x.cosx.=2cos4x.cosx`
`therefore cot4x(sin5x+sin3x)=2cos4xcosx`………………..(1)
`=(cosx)/(sinx)(2.cos(5x+3x)/(2)sin(5x-3x)(2))`
`=(cosx)/(sinx). 2cos4x.sinx=2cos4xcosx`
`therefore cotx(sinx-sin3x)=2cos4xcosx`…………(2)
From equations, (1) and (2),
`cot4x(sin5x+sin3x)=cotx(sin5x-sin3x)`
LHS =RHS Hence proved.