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 agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
prove that \( 3 sin \dfrac{π}{6} sec \dfrac{π}{3} – 4 sin \dfrac{5π}{6} cot \dfrac{π}{4} = 1\)
prove that \( tan \dfrac{11π}{3} – 2 sin \dfrac{4π}{6} – \dfrac{3}{4} cosec^2 \dfrac{π}{4} + 4 cos^2 \dfrac{17π}{6} = \dfrac{(3 – 4\sqrt{3})}{2}\)
prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0
prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0