Consider the following diagram:
![enter image source here]()
The diagram is just to say that we know the two legs of the triangle, which lie in quadrant four. In other words, we can already state tanalphatanα and cotalphacotα, but to find the other four trigonometric functions we need to know the length of the hypotenuse, or rr.
We can do this by pythagorean theorem:
r^2 = (-5)^2 + 9^2r2=(−5)2+92
r = sqrt(25 + 81)r=√25+81
r = sqrt(106)r=√106
Now, we can state the value of all 6 trigonometric ratios.
sinalpha = "opposite"/"hypotenuse" = -5/sqrt(106) = (-5sqrt(106))/106sinα=oppositehypotenuse=−5√106=−5√106106
cscalpha = 1/("opposite"/"hypotenuse") = 1/(-5/(sqrt(106))) = -sqrt(106)/5cscα=1oppositehypotenuse=1−5√106=−√1065
cosalpha = "adjacent/hypotenuse" = 9/sqrt(106) = (9sqrt(106))/106cosα=adjacent/hypotenuse=9√106=9√106106
secalpha = 1/("adjacent"/"hypotenuse") = 1/(9/sqrt(106)) = sqrt(106)/9secα=1adjacenthypotenuse=19√106=√1069
tanalpha = "opposite"/"adjacent" = -5/9tanα=oppositeadjacent=−59
cotalpha = 1/("opposite"/"adjacent") = -9/5cotα=1oppositeadjacent=−95
Hopefully this helps!
