Question

Find the values of other five trigonometric functions if $\cot x=\frac{3}{4},x$ lies in third quadrant.

Answer 1

Step 1: Solve for value of $\text{cosec}x$
$\cot x=\frac{3}{4}$ and $x$ lies in third quadrant
We know that $1+{\cot }^{2}x={\text{cosec}}^{2}x$
$\Rightarrow {\text{cosec}}^{2}x=1+{\left(\frac{3}{4}\right)}^2$
$\Rightarrow {\text{cosec}}^{2}x=1+\frac{9}{16}$
$\Rightarrow {\text{cosec}}^{2}x=\frac{25}{16}$
$\Rightarrow \text{cosec}x=\pm \frac{5}{4}$
As $x$ is in third quadrant, $\text{cosec}x$ is negative.
$\therefore \text{cosec}x=-\frac{5}{4}$

Step 2: Solve for value of $\sin x$
We know that $\sin x=\frac{1}{\text{cosec}x}$
$\therefore \sin x=\frac{1}{-\frac{5}{4}}=-\frac{4}{5}$

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

Step 4: Solve for value of $\cos x$
We know that $\tan x=\frac{\sin x}{\cos x}$
$\therefore \cos x=\frac{\sin x}{\tan x}=\frac{-\frac{4}{5}}{\frac{4}{3}}=-\frac{3}{5}$

Step 5: Solve for value of $\sec x$
We know that $\sec x=\frac{1}{\cos x}$
$\therefore \sec x=\frac{1}{-\frac{3}{5}}=-\frac{5}{3}$