Here given that,
\( tanx=\sqrt{3}\)
We know that
\( tan\dfrac{\pi}{3}=\sqrt{3}\)
and \( tan\dfrac{4\pi}{3}= tan(\pi+\dfrac{\pi}{3})\)
\( = tan\dfrac{\pi}{3}=\sqrt{3}\)
Therefore, the principal solutions are
\(x= \dfrac{\pi}{3}\) and \( \dfrac{4\pi}{3}\)
Now, \( tanx=tan\dfrac{\pi}{3}\)
Which implies,
\( x=n\pi + \dfrac{\pi}{3},n\in Z\)
Answered by Abhisek | 1 year agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
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