Question

Find the value of other five trigonometric ratios:

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

Answer 1

$\cos x=-\frac{1}{2}$
$\therefore \sec x=\frac{1}{\cos x}=-2$
Now $\because {\sin }^{2}x+{\cos }^{2}x=1$
$⟹{\sin }^{2}x=1-{\cos }^{2}x⟹{\sin }^{2}x=1-\frac{1}{4}=\frac{3}{4}⟹\sin x=\pm \frac{\sqrt{3}}{2}$
Since $x$ lies in third quadrant, the value of $\sin x$ will be negative.
$\therefore \sin x=-\frac{\sqrt{3}}{2}$

$\csc x=\frac{1}{\sin x}=-\frac{2}{\sqrt{3}}$
$\tan x=\frac{\sin x}{\cos x}=\frac{(-\frac{\sqrt{3}}{2})}{(-\frac{1}{2})}=\sqrt{3}$
$\cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}}$