Substituting the values of \( cot\dfrac{\pi}{6},cosec \dfrac{5\pi}{6},tan\dfrac{\pi}{6}\) on left hand side,
L.H.S. = \((\sqrt{3})^2+cosec\dfrac{5\pi}{6}+3tan^2\dfrac{\pi}{6}\)
\( (\sqrt{3})^2+cosec(\pi-\dfrac{\pi}{6})+3(\dfrac{1}{\sqrt{3}})^2\)
\( 3+cosec\dfrac{\pi}{6}+3\times \dfrac{1}{3}\)
Since cosecx repeat its value after interval of 2π,
We have, \( cosec\dfrac{5\pi}{6}=cosec\dfrac{\pi}{6}\)
L.H.S= 3+ 2 +1 =1
= 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