Question

Find the value of other five trigonometric ratios:
$\sec x = \dfrac{{13}}{5}$ , x lies in the fourth quadrant.

Answer 1

Hint: Given that x lies in fourth quadrant. Now if x lies in the fourth quadrant, only \[\cos x\] and \[\sec x\] is positive.
Also, note the following important formulae:
$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
${\sin ^2}x + {\cos ^2}x = 1$
\[{\sec ^2}x - {\tan ^2}x = 1\]
\[{\operatorname{cosec} ^2}x - {\cot ^2}x = 1\]
Now, the value of \[\sec x\] is given. Therefore find the value of the other five trigonometric ratios with the help of aforementioned formulae.

Complete step-by-step answer:
Given, $\sec x = \dfrac{{13}}{5}$
Therefore $\cos x = \dfrac{1}{{\sec x}} = \dfrac{5}{{13}}$
Now,
${\sin ^2}x + {\cos ^2}x = 1$
$ \Rightarrow {\sin ^2}x = 1 - {\cos ^2}x$
Taking square root we get,
$ \Rightarrow \sin x = \pm \sqrt {1 - {{\cos }^2}x} $
On substituting the value of \[\cos x\] we get,
$ \Rightarrow \sin x = \pm \sqrt {1 - {{\left( {\dfrac{5}{{13}}} \right)}^2}} $
On simplification we get,
$ \Rightarrow \sin x = \pm \sqrt {1 - \dfrac{{25}}{{169}}} = \pm \sqrt {\dfrac{{144}}{{169}}} $
As, x lies in the fourth quadrant, so value of \[\sin x\] will be negative,
$ \Rightarrow \sin x = - \dfrac{{12}}{{13}}{\text{, }}$
Therefore, $\operatorname{cosecx} = \dfrac{1}{{\sin x}} = - \dfrac{{13}}{{12}}$
Again,
\[\because {\sec ^2}x - {\tan ^2}x = 1\]
 \[ \Rightarrow {\tan ^2}x = {\sec ^2}x - 1\]
On taking square root we get,
\[ \Rightarrow \tan x = \pm \sqrt {{{\sec }^2}x - 1} \]
On substituting the value of \[\sec x\] we get,
\[ \Rightarrow \tan x = \pm \sqrt {{{\left( {\dfrac{{13}}{5}} \right)}^2} - 1} \]
On simplification we get,
\[ \Rightarrow \tan x = \pm \sqrt {\dfrac{{169}}{{25}} - 1} = \pm \sqrt {\dfrac{{144}}{{25}}} \]
As, x lies in the fourth quadrant, so value of \[\tan x\] will be negative,
 \[ \Rightarrow \tan x = - \dfrac{{12}}{5}{\text{ }}\]
Therefore $\cot x = \dfrac{1}{{\tan x}} = - \dfrac{5}{{12}}$
Hence, when $\sec x = \dfrac{{13}}{5}$ and x lies in fourth quadrant, the other five trigonometric ratios are :
$\cos x = \dfrac{5}{{13}}$, $\sin x = - \dfrac{{12}}{{13}}$ , $\operatorname{cosecx} = - \dfrac{{13}}{{12}}$, \[\tan x = - \dfrac{{12}}{5}\]and $\cot x = - \dfrac{5}{{12}}$


Note: Note the following important formulae:
1.$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
2.${\sin ^2}x + {\cos ^2}x = 1$
3.\[{\sec ^2}x - {\tan ^2}x = 1\]
4.\[{\operatorname{cosec} ^2}x - {\cot ^2}x = 1\]
5.$\sin ( - x) = - \sin x$
6.$\cos ( - x) = \cos x$
7.$\tan ( - x) = - \tan x$
8.$\sin \left( {2n\pi \pm x} \right) = \sin x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
9.$\cos \left( {2n\pi \pm x} \right) = \cos x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
10.$\tan \left( {n\pi \pm x} \right) = \tan x{\text{ , period }}\pi {\text{ or 18}}{0^ \circ }$
Sign convention: