Question

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Answer 1

To prove ---->

$\dfrac{cos9x - cos5x}{sin17x - sin3x} = - \dfrac{sin2x}{cos10x}$

Concept use---->

1)

$cos \alpha - cos \beta = - 2sin( \dfrac{ \alpha + \beta }{2} ) \: sin( \dfrac{ \alpha - \beta }{2}$

2)

$sin \alpha - sin \beta = 2 \: cos( \dfrac{ \alpha + \beta }{2}) \: sin( \dfrac{ \alpha - \beta }{2})$

Proof ----> LHS

$= \dfrac{cos9x - cos5x}{sin17x - sin3x}$

$= \dfrac{ - 2 \: sin( \dfrac{9x + 5x}{2}) \: sin( \dfrac{9x - 5x}{2}) }{2 \: cos( \dfrac{17x + 3x}{2}) \: sin( \dfrac{17x - 3x}{2}) }$

$2 \: is \: cancel \: out \: from \: numerator \: and \: denominator$

$= - \dfrac{sin( \dfrac{14x}{2}) \: sin( \dfrac{4x}{2}) }{cos( \dfrac{20x}{2}) \: sin( \dfrac{14x}{2}) }$

$= - \dfrac{sin7x \: \: \: sin2x}{cos10x \: \: \: sin7x} \\ \\ sin7x \: cancel \: out \: from \: numerator \: and \: denominator \\ = - \dfrac{sin2x}{cos10x}$

= RHS

Answer 2

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