Question

Question 6 Prove that: [ (sin 7x + sin 5x) + (sin 9x + sin 3x) ] / [ (cos 7x + cos 5x) + (cos 9x + cos 3x) ] = tan 6x

Class X1 - Maths -Trigonometric Functions Page 82

Answer 1

LHS = [(sin7x + sin5x) + (sin9x + sin3x)]/[(cos7x + cos5x) + (cos9x + cos3x)]

use the formula,
sinC + sinD = 2sin(C+D)/2.cos(C-D)/2
cosC + cosD = 2cos(C+D)/2.cos(C-D)/2

= {2sin6x.cosx +2sin6x.cos3x}/{2cos6x.cosx + 2cos6x.cos3x}

= 2sin6x.(cosx + cos3x)/2cos6x(cosx+cos3x)

= sin6x/cos6x

= tan6x = RHS

Answer 2

Answer:

LHS = [(sin7x + sin5x) + (sin9x + sin3x)]/[(cos7x + cos5x) + (cos9x + cos3x)]

use the formula,

sinC + sinD = 2sin(C+D)/2.cos(C-D)/2

cosC + cosD = 2cos(C+D)/2.cos(C-D)/2

= {2sin6x.cosx +2sin6x.cos3x}/{2cos6x.cosx + 2cos6x.cos3x}

= 2sin6x.(cosx + cos3x)/2cos6x(cosx+cos3x)

= sin6x/cos6x

= tan6x = RHS