Substituting the values of \( sin\dfrac{\pi}{6},cos \dfrac{\pi}{6},tan\dfrac{\pi}{4}\) on left hand side,
\( sin^2\dfrac{\pi}{6}+cos^2 \dfrac{\pi}{6}-tan^2\dfrac{\pi}{4}\)
\(( \dfrac{1}{2})^2+(\dfrac{1}{2})^2-(1)^2\)
\( \dfrac{1}{4}+\dfrac{1}{4}-1\)
= \(- \dfrac{1}{2}= R.H.S.\)
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