Question

If $\cos x=\frac{-1}{2}$ and $x$ lies in third quadrant, then

• A. $\tan x=-\sqrt{3}$ and $\cot x=\frac{-1}{\sqrt{3}}$
• B. $\sec x=2$ and $\mathrm{cosec}x=\frac{2}{\sqrt{3}}$
• C. $\sin x=\frac{-\sqrt{3}}{2}$
• D. All are correct

Answer 1

Answer: (C)

$\cos x=-\frac{1}{2}$
Given that $x$ lies in third quadrant.
i.e., $\pi <x<\frac{3\pi }{2}$
We have, ${\sin }^{2}x+{\cos }^{2}x=1\Rightarrow {\sin }^{2}x=1-{\cos }^{2}x$
$\Rightarrow {\sin }^{2}x=1-{\left(-\frac{1}{2}\right)}^2$
$\Rightarrow {\sin }^{2}x=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$
$\because$ In third quadrant $\sin x$ is negative, so we will leave its positive value.
i.e., $\sin x=-\frac{\sqrt{3}}{2}$
Now, $\tan x=\frac{\sin x}{\cos x}=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}=\sqrt{3}$
$\cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}}$
and $\sec x=\frac{1}{\cos x}=-2,$
$\mathrm{cosec}x=\frac{1}{\sin x}=-\frac{2}{\sqrt{3}}$