Answer:
Given : The point with coordinates (6,-5) lies on its terminal side.
To find : The values of the six trigonometric functions of an angle in standard position
Solution :
The coordinates (6,-5) lies in the 4 quadrant.
Refer the attached figure.
where x=6 is the base
y= -5 is the perpendicular
Applying Pythagoras theorem,
[tex]H^2=P^2+B^2[/tex]
[tex]H^2=(-5)^2+6^2[/tex]
[tex]H^2=25+36[/tex]
[tex]H=\sqrt{61}[/tex]
Now, finding trigonometric function
Note - In fourth quadrant only cos and sec is positive rest are negative.
1 ) [tex]\sin x= - \frac{P}{H}[/tex]
[tex]\sin x= - \frac{-5}{\sqrt{61}}[/tex]
[tex]\sin x= \frac{5}{\sqrt{61}}[/tex]
2) [tex]\csc x=\frac{1}{\sin x}[/tex]
[tex]\csc x=\frac{1}{\frac{5}{\sqrt{61}}}[/tex]
[tex]\csc x=\frac{\sqrt{61}}{5}[/tex]
3) [tex]\cos x= \frac{B}{H}[/tex]
[tex]\cos x=\frac{6}{\sqrt{61}}[/tex]
4) [tex]\sec x=\frac{1}{\cos x}[/tex]
[tex]\sec x=\frac{1}{\frac{6}{\sqrt{61}}}[/tex]
[tex]\sec x=\frac{\sqrt{61}}{6}[/tex]
5) [tex]\tan x= - \frac{P}{B}[/tex]
[tex]\tan x= - \frac{-5}{6}[/tex]
[tex]\tan x= \frac{5}{6}[/tex]
6) [tex]\cot x=\frac{1}{\tan x}[/tex]
[tex]\cot x=\frac{1}{\frac{5}{6}}[/tex]
[tex]\cot x=\frac{6}{5}[/tex]
