# Find the values of other five trigonometric functions if sec x = 13/5

Question:

Find the values of other five trigonometric functions if $\sec x=\frac{13}{5}, x$ lies in fourth quadrant

Solution:

$\sec x=\frac{13}{5}$

$\cos x=\frac{1}{\sec x}=\frac{1}{\left(\frac{13}{5}\right)}=\frac{5}{13}$

$\sin ^{2} x+\cos ^{2} x=1$

$\Rightarrow \sin ^{2} x=1-\cos ^{2} x$

$\Rightarrow \sin ^{2} x=1-\left(\frac{5}{13}\right)^{2}$

$\Rightarrow \sin ^{2} x=1-\frac{25}{169}=\frac{144}{169}$

$\Rightarrow \sin x=\pm \frac{12}{13}$

Since $x$ lies in the $4^{\text {th }}$ quadrant, the value of $\sin x$ will be negative.

$\therefore \sin x=-\frac{12}{13}$

$\operatorname{cosec} x=\frac{1}{\sin x}=\frac{1}{\left(-\frac{12}{13}\right)}=-\frac{13}{12}$

$\tan x=\frac{\sin x}{\cos x}=\frac{\left(\frac{-12}{13}\right)}{\left(\frac{5}{13}\right)}=-\frac{12}{5}$

$\cot x=\frac{1}{\tan x}=\frac{1}{\left(-\frac{12}{5}\right)}=-\frac{5}{12}$