Question

Find the value of other five trigonometric ratios:

$\sec x=\frac{13}{5}$, x lies in fourth quadrant.

Answer 1

$\sec x=\frac{13}{5}$
$\Rightarrow \cos x=\frac{1}{\sec x}=\frac{5}{13}$
Now $\because {\sin }^{2}x+{\cos }^{2}x=1$
$⟹{\sin }^{2}x=1-{\cos }^{2}x=1-\frac{25}{169}=\frac{144}{169}$
$⟹\sin x=\pm \frac{12}{13}$
Since $x$ lies in fourth quadrant, the value of $\sin x$ will be negative.
$\therefore \sin x=-\frac{12}{13}$
$\csc x=\frac{1}{\sin x}=-\frac{13}{12}$
$\tan x=\frac{\sin x}{\cos x}=\frac{(-\frac{12}{13})}{\left(\frac{5}{13}\right)}=-\frac{12}{5}$
$\cot x=\frac{1}{\tan x}=-\frac{5}{12}$