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: $\frac{1}{\sqrt{3}}$
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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