Question

Find the value of other five trigonometric ratios:

$\cot x=\frac{3}{4}$, x lies in third quadrant.

Answer 1

$\cot x=\frac{3}{4}$
$\Rightarrow \tan x=\frac{1}{\cot x}=\frac{4}{3}$
Now $\because 1+{\tan }^{2}x={\sec }^{2}x$
$⟹{\sec }^{2}x=1+\frac{16}{9}⟹{\sec }^{2}x=\frac{25}{9}$
$\sec x=\pm \frac{5}{3}$
Since $x$ lies in third quadrant, the value of $\sec x$ will be negative.
$\therefore \sec x=-\frac{5}{3}$
$\cos x=\frac{1}{\sec x}=-\frac{3}{5}$
$\tan x=\frac{\sin x}{\cos x}$$⟹\sin x=\left(\frac{4}{3}\right)\times (-\frac{3}{5})=-\frac{4}{5}$
$\therefore \csc x=\frac{1}{\sin x}=-\frac{5}{4}$