Prove the following: (sin x - sin 3x) / (sin²x - cos²x) = 2sin x

Solution:

LHS = (sin x - sin 3x) / (sin²x - cos²x)

= [2cos (x + 3x)/2 sin (x - 3x)/2] / -[cos²x - sin²x]

[Since sin A - sin B = 2 cos [(A + B) / 2] sin [(A - B) / 2]]

= 2cos 2x sin (-x) / -cos 2x 

[By double angle formulas, cos 2x = cos²x - sin²x ]

= 2cos 2x (-sin x) / (- cos 2x) 

[As sin (-A) = - sin A]

= 2sin x

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 20

Prove the following: (sin x - sin 3x) / (sin²x - cos²x) = 2sin x

Summary:

We got, (sin x - sin 3x) / (sin²x - cos²x) = 2sin x. Hence Proved