Question

prove that cos9x-cos5x/sin17x-sin3x=-sin2x/cos10x

Answer 1

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Answer 2

$\underline{\underline{\sf{\red{Given\::-}}}}$

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

$\underline{\underline{\sf{\red{To\:Prove\::-}}}}$

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

$\underline{\underline{\sf{\red{Proof\::-}}}}$

⌬ Identities that we will use :-

$\sf cos\:A - cos\:B = -2sin\bigg\lgroup \dfrac{A + B}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{A - B}{2} \bigg\rgroup$

$\sf sin\:A - sin\:B = 2cos\bigg\lgroup \dfrac{A + B}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{A - B}{2} \bigg\rgroup$

Now,

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

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin\bigg\lgroup \dfrac{9x + 5x}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{9x - 5x}{2} \bigg\rgroup}{2cos\bigg\lgroup \dfrac{17x + 3x}{2} \bigg\rgroup\:.\: sin \bigg\lgroup \dfrac{17x - 3x}{2} \bigg\rgroup}$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin\bigg\lgroup \dfrac{14x}{2} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{4x}{2} \bigg\rgroup}{2cos\bigg\lgroup \dfrac{20x}{2} \bigg\rgroup\:.\: sin \bigg\lgroup \dfrac{14x}{2} \bigg\rgroup}$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin\bigg\lgroup \dfrac{\cancel{14}x}{\cancel{2}} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{\cancel{4}x}{\cancel{2}} \bigg\rgroup}{2cos\bigg\lgroup \dfrac{\cancel{20}x}{\cancel{2}} \bigg\rgroup\:.\:sin \bigg\lgroup \dfrac{\cancel{14}x}{\cancel{2}} \bigg\rgroup}\:(Cancelling)$

$\longrightarrow\:\sf L.H.S = \dfrac{-2sin7x\:.\:sin2x}{2cos10x\:.\:sin7x}$

$\longrightarrow\:\sf L.H.S = \dfrac{-\cancel{2}\cancel{sin7x}\:.\:sin2x}{\cancel{2}cos10x\:.\:\cancel{sin7x}}\:(Cancelling$

$\longrightarrow\:\sf L.H.S = \dfrac{-sin2x}{cos10x}$

$\pink{\bigstar}\:\underline{\underline{\bf{\blue{L.H.S = R H.S}}}}$

$\:\:\qquad\qquad\therefore\:\underline{\underline{\sf{\red{Hence,\:Proved!}}}}$