The given trigonometric equation is,
cot4x( sin5x+sin3x )=cotx( sin5x−sin3x )
Solve left hand side,
L.H.S.=cot4x( sin5x+sin3x ) =cot4x×2sin ( 5x+3x ) 2 cos ( 5x−3x ) 2 =cot4x×2sin 8x 2 cos 2x 2 = cos4x sin4x ×2sin4xcosx
Further simplify,
L.H.S.=2cos4xcosx =2cos4xsinx× cosx sinx =[ { sin( 4x+x )−sin( 4x−x ) } ]×cotx =cotx( sin5x−sin3x )
Right hand side is,
R.H.S.=cotx( sin5x−sin3x )
Hence, L.H.S.=R.H.S.
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)