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

Asked by Pragya Singh | 1 year ago |  83

Solution :-

We know that,

cosA-cosB=-2sin $$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$

sinA-sinB=2cos$$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$

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

$$L.H.S.= \dfrac{cos9x-cos5x}{sin17x-sin3x}$$

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

$$\dfrac{-2sin7x.sin2x}{2cos10x.sin7x}$$

$$- \dfrac{sin2x}{cos10x}$$=  R.H.S.

Hence proved.

Answered by Abhisek | 1 year ago

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