Question 16 Prove that [cos(9x) - cos(5x)] / [sin(17x) - sin(3x)] = - sin 2x / cos 10x
Class X1 - Maths -Trigonometric Functions Page 73
Answers
Answered by
60
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
Answered by
11
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
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