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Prove the following: sinx-sin3xsin2x-cos2x= 2sinx

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Question

Prove the following:

`(sin x - sin 3x)/(sin^2 x - cos^2 x) =  2sin x`

Sum
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Solution

We have, L.H.S. = `(sin x - sin 3x)/(sin^2 x - cos^2 x)`

= `(2sin ((x - 3x)/2) cos ((x + 3x)/2))/(-cos^2x - sin^2x)` 

= `(2sin (-x) cos (2x))/(-cos2x)`

= `(- 2sin x cos(2x))/(-cos2x)`  [∵ cos2x - sin2x = cos 2x]

= `(-2 sinx)/-1`  [∵ cos2x - sin2x = cos 2x]

= 2 sin x = R.H.S.

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Chapter 3: Trigonometric Functions - EXERCISE 3.3 [Page 67]

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NCERT Mathematics [English] Class 11
Chapter 3 Trigonometric Functions
EXERCISE 3.3 | Q 20. | Page 67

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