Here given that, \( cosx=-\dfrac{1}{2}\)
Therefore we have,
\( secx=\dfrac{1}{cosx}\)
= \( \dfrac{1}{(-\dfrac{1}{2})}\) = -2
We know that
sin2 x + cos2 x = 1
We can write it as
sin2 x = 1 – cos2 x
Substituting cosx = \(- \dfrac{1}{2}\) in the formula, we obtain,
\( sin^2x=1-(-\dfrac{1}{2})^2\)
\( sin^2x=1-\dfrac{1}{4}\) = \(\dfrac{3}{4}\)
\( sinx\pm \dfrac{\sqrt{3}}{2}\)
Since x lies in the 3rd quadrant, the value of sin x will be negative.
\( sinx=- \dfrac{\sqrt{3}}{2}\)
Therefore, \( cosecx=\dfrac{1}{sinx}\)
= \( \dfrac{1}{(\dfrac{-\sqrt{3}}{2})}=-\dfrac{2}{\sqrt{3}}\)
Hence,
\( tanx=\dfrac{sinx}{cosx}\)
\( \dfrac{(-\dfrac{\sqrt{3})}{2}}{(-\dfrac{1}{2})}=\sqrt{3}\)
And
\( cotx=\dfrac{1}{tanx}\)
= \(\dfrac{1}{\sqrt{3}}\)
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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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
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