\frac { \operatorname { sin } \theta + \operatorname { sin } 3 \theta + \operatorname { sin } 5 \theta + \operatorname { sin } 7 \theta } { \operatorname { cos } \theta + \operatorname { cos } 3 \theta + \operatorname { cos } 5 \theta + \operatorname { cos } 7 \theta } = \operatorname { tan } 4 \theta

\frac { \operatorname { sin } \theta + \operatorname { sin } 3 \theta

1.	\frac { \operatorname { sin } \theta + \operatorname { sin } 3 \theta + \operatorname { sin } 5 \theta + \operatorname { sin } 7 \theta } { \operatorname { cos } \theta + \operatorname { cos } 3 \theta + \operatorname { cos } 5 \theta + \operatorname { cos } 7 \theta } = \operatorname { tan } 4 \theta
Answer» Taking the numerator i.e. sinx + sin3x+sin5x +sin7x= (sinx +sin7x) + (sin3x + sin5x)applying formula of sinc +sind , we get ,= 2sin4xcos3x + 2sin4xcosx =2sin4x (cos3x+cosx)--- -------- ---- --------1Similarly taking the denominator i.e. (cosx + cos7x) + (cos3x + cos5x)applying formula of cosC +cosD, we get = 2cos4xcos3x + 2cos4xcosx =2cos4x (cos3x + cosx) ---------- ------------2Now dividing 1 by 2 =we get sin4x/cos4x = tan4x