Question : If $\cos x=-\frac{1}{2}, x$ lies in third quadrant, then $\tan x=?$
Option 1: $\sqrt{3}$
Option 2: $\frac{\sqrt{3}}{2}$
Option 3: $\frac{2}{\sqrt{3}}$
Option 4: $\frac{1}{\sqrt{3}}$
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Correct Answer: $\sqrt{3}$
Solution : Given: $\cos x=-\frac{1}{2}, x$ lies in third quadrant. $\sin^2x=1-\cos^2x=1-(-\frac{1}{2})^2=\frac{3}{4}$ ⇒ $\sin x= \pm \frac{\sqrt{3}}{2}$ As $x$ is in the 3rd quadrant therefore $\sin x=- \frac{\sqrt{3}}{2}$ $\therefore \tan x=\frac{\sin x}{\cos x}=\frac{- \frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}$ Hence, the correct answer is $\sqrt{3}$.
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