Question

Find the values of other five trigonometric functions if $\cos x=-\frac{1}{2},x$ lies in third quadrant

Answer 1

Step 1: Solve for value of $\sin x$
We know that ${\sin }^{2}x+{\cos }^{2}x=1$
$\Rightarrow {\sin }^{2}x+{\left(\frac{-1}{2}\right)}^2=1$
$\Rightarrow {\sin }^{2}x+\frac{1}{4}=1$
$\Rightarrow {\sin }^{2}x=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$
Since $x$ is in third quadrant, $\sin x$ is negative.
$\therefore \sin x=-\frac{\sqrt{3}}{2}$

Step 2: Solve for value of $\text{cosec}x$
We know, $\text{cosec}x=\frac{1}{\sin x}$
$\therefore \text{cosec}x=\frac{1}{\frac{-\sqrt{3}}{2}}=\frac{-2}{\sqrt{3}}$

Step 3: Solve for value of $\tan x$
We know that $\tan x=\frac{\sin x}{\cos x}$
$\therefore \tan x=\frac{(-\frac{\sqrt{3}}{2})}{\left(\frac{-1}{2}\right)}=\sqrt{3}$

Step 4: Solve for value of $\cot x$
We know that $\cot x=\frac{1}{\tan x}$
$\therefore \cot x=\frac{1}{\sqrt{3}}$

Step 5: Solve for value of $\sec x$
We know that $\sec x=\frac{1}{\cos x}$
$\therefore \sec x=\frac{1}{(-\frac{1}{2})}=-2$