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Prove the following: cos9x-cos5xsin17x-sin3x=-sin2xcos10x

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Question

Prove the following:

`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`

Sum
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Solution

we know that,

cos A - cos B = -2 sin `((A + B)/2) sin((A - B)/2), sin A - sin B = 2 cos ((A + B)/2) sin ((A - B)/2)`

L.H.S. = `(cos9x - cos5x)/(sin17x - sin 3x)`

= `(-2sin  ((9x + 5x)/2) sin ((9x - 5x)/2))/(2cos((17x+3x)/2) sin((17x - 3x)/2))`

= `(-2sin7xsin2x)/(2cos10xsin7x)`

= `(-sin2x)/(cos10x)` = 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 16. | Page 67

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