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

Question

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

Aicharya · Accepted Answer

$\cot 	heta=\sqrt{3}$ and $\sec 	heta=\frac{-2}{\sqrt{3}}$ i.e., $	an 	heta=\frac{1}{\sqrt{3}}$ and $\cos 	heta=-\frac{\sqrt{3}}{2}$ $	herefore \quad 	heta$ lies in $3^{	ext {rd }}$ quadrant $	herefore \quad 	heta=\frac{7 \pi}{6}$

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

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