\text { prove that }\frac { \cos 9 x - \cos 5 x } { \sin 17 x - \sin 3

1.	\text { prove that }\frac { \cos 9 x - \cos 5 x } { \sin 17 x - \sin 3 x } = - \frac { \sin 2 x } { \cos 10 x }
Answer» LHS = (cos9x - cos5x)/(sin17x-sin3x) Use the formula,cosC - cosD = 2sin(C + D)/2.sin(D-C)/2 sinC-sinD = 2cos(C+D)/2.sin(C-D)/2 = {2sin(9x+5x)/2.sin(5x-9x)/2}/{2cos(17x+3x)/2.sin(17x-3x)/2}=-(sin7x.sin2x)/(cos10x.sin7x)= - sin2x/cos10x = RHS

\text { prove that }\frac { \cos 9 x - \cos 5 x } { \sin 17 x - \sin 3 x } = - \frac { \sin 2 x } { \cos 10 x }

Answer»

LHS = (cos9x - cos5x)/(sin17x-sin3x) Use the formula,cosC - cosD = 2sin(C + D)/2.sin(D-C)/2 sinC-sinD = 2cos(C+D)/2.sin(C-D)/2

= {2sin(9x+5x)/2.sin(5x-9x)/2}/{2cos(17x+3x)/2.sin(17x-3x)/2}=-(sin7x.sin2x)/(cos10x.sin7x)= - sin2x/cos10x = RHS