If tan theta = - frac{1}{{sqrt 3 }} and sin t | vedclass

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(b) $	an 	heta = \frac{{ - 1}}{{\sqrt 3 }} = 	an \left( {\pi - \frac{\pi }{6}} ight)$,

$\sin 	heta = \frac{1}{2} = \sin \left( {\pi - \frac{\

Vedclass · Accepted Answer

$\frac{{5\pi }}{6}$

Vedclass · Answer

(b) $	an 	heta = \frac{{ - 1}}{{\sqrt 3 }} = 	an \left( {\pi - \frac{\pi }{6}} ight)$,

$\sin 	heta = \frac{1}{2} = \sin \left( {\pi - \frac{\pi }{6}} ight)$

and $\cos 	heta = \frac{{ - \sqrt 3 }}{2}$= $\cos \left( {\pi - \frac{\pi }{6}} ight)$

Hence principal value is $	heta = \frac{{5\pi }}{6}$.

If $\tan \theta = - \frac{1}{{\sqrt 3 }}$ and $\sin \theta = \frac{1}{2}$, $\cos \theta = - \frac{{\sqrt 3 }}{2}$, then the principal value of $\theta $ will be

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