Here it is given that,
\( cotx=-\sqrt{3}\)
Now we know that
\( cot=\dfrac{\pi}{6}=\sqrt{3}\)
And
\( cot=(\pi-\dfrac{\pi}{6})\)
= \(- cot=\dfrac{\pi}{6}\)
= \( -\sqrt{3}\)
\( cot=(2\pi-\dfrac{\pi}{6})=-cot\dfrac{\pi}{6}\)
= \( -\sqrt{3}\)
Therefore we have,
\( cot\dfrac{5\pi}{6}= -\sqrt{3}\) and \( cot\dfrac{11\pi}{6}= -\sqrt{3}\)
Therefore, the principal solutions are
\(x=\dfrac{5\pi}{6}\;and\;\dfrac{11\pi}{6}\)
And we know \(cotx= \dfrac{1}{tanx}\)
Therefore we have,
\( tanx=tan\dfrac{5\pi}{6}\)
Which implies,
\( x=n\pi \pm \dfrac{5\pi}{6}\) , where \( n\in Z \)
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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