Question

Question 16 Prove that [cos(9x) - cos(5x)] / [sin(17x) - sin(3x)] = - sin 2x / cos 10x

Class X1 - Maths -Trigonometric Functions Page 73

Answer 1

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

Answer 2

Answer:

LHS = RHS

Step-by-step explanation:

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