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

Solution:

LHS = [(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)]

= [2sin {(7x + 5x) / 2} cos {(7x - 5x) / 2} + 2sin {(9x + 3x) / 2} cos {(9x - 3x) / 2}] / [2cos {(7x + 5x) / 2} cos {(7x - 5x) / 2} + 2cos {(9x + 3x) / 2} cos {(9x - 3x) / 2}]
[Because sin A + sin B = 2sin {(A + B) / 2} cos {(A - B) / 2} and cos A + cos B = 2cos {(A + B) / 2} cos {(A - B) / 2}]

= [2sin 6x cos x + 2sin 6x cos 3x] / [2cos 6x cos x + 2cos 6x + 2cos 6x cos 3x]

= [2sin 6x(cos x + cos 3x)] / [2cos 6x(cos x + cos 3x)]

= sin 6x / cos 6x

= tan 6x 

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise ME Question 6

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

Summary:

We got, ([(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)] = tan 6x. Hence Proved