Question

prove that cos(9x)-cos(5x)/sin(17x)-sin(3x)=-sin(2x)/cos(10x)​

Answer 1

Given to prove that :-

$\dfrac{cos(9x) - cos(5x)}{sin(17x) - sin(3x)} = \dfrac{-sin(2x)}{cos(10x)}$

SOLUTION :-

Take L.H.S

$\dfrac{cos(9x) - cos(5x)}{sin(17x) - sin(3x)}$

Numerator is in form of cosC-cosD

Denominator is in form of sinC - sinD

So,

${cosC-cosD} = -2 sin\bigg(\dfrac{C+D}{2}\bigg) sin\bigg(\dfrac{C-D}{2}\bigg)$

${sinC-sinD} = 2 cos\bigg(\dfrac{C+D}{2}\bigg) sin\bigg(\dfrac{C-D}{2}\bigg)$

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

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

$\bigg(\dfrac{ - 2sin7x \: sin2x}{2cos10x \: sin7x}\bigg)$

$\bigg(\dfrac{-sin7x \: sin2x}{cos10x \: sin7x}\bigg)$

$\dfrac{ - sin2x}{cos10x}$

Hence proved !!

L.H.S = R.H. S