If angle theta in [0, 2pi] satisfies both the | vedclass

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(B) Given equations are $\cot 	heta = \sqrt{3}$ and $\sqrt{3} \sec 	heta + 2 = 0$. From $\cot 	heta = \sqrt{3}$,we have $	an 	heta = \frac{1}{\sq

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$\frac{7 \pi}{6}$

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(B) Given equations are $\cot 	heta = \sqrt{3}$ and $\sqrt{3} \sec 	heta + 2 = 0$. From $\cot 	heta = \sqrt{3}$,we have $	an 	heta = \frac{1}{\sqrt{3}}$. This implies $	heta$ is in the $1^{	ext{st}}$ or $3^{	ext{rd}}$ quadrant. From $\sqrt{3} \sec 	heta + 2 = 0$,we have $\sec 	heta = -\frac{2}{\sqrt{3}}$,which means $\cos 	heta = -\frac{\sqrt{3}}{2}$. Since $\cos 	heta$ is negative,$	heta$ must be in the $2^{	ext{nd}}$ or $3^{	ext{rd}}$ quadrant. Combining both conditions,$	heta$ must lie in the $3^{	ext{rd}}$ quadrant. In the $3^{	ext{rd}}$ quadrant,$	an 	heta = \frac{1}{\sqrt{3}}$ corresponds to $	heta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.

If angle $\theta$ in $[0, 2\pi]$ satisfies both the equations $\cot \theta = \sqrt{3}$ and $\sqrt{3} \sec \theta + 2 = 0$,then $\theta$ is equal to

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