if cos θ \theta θ >0 and tan θ \theta θ <0 then this falls on the fourth quadrant. The sin will also be negative.
let x be the adjacent, y the opposite and r the hypotenuse of a right angled triangle where θ \theta θ is the angle formed as x and r intersect.
c o s θ = x r cos \theta=\frac x r cos θ = r x
x=1, r=3
y = r 2 − x 2 y=\sqrt {r^2-x^2} y = r 2 − x 2
y = 3 2 − 1 2 = 8 y=\sqrt{3^2-1^2}=\sqrt 8 y = 3 2 − 1 2 = 8
s e c θ = 1 c o s θ = 1 1 3 = 3 sec \theta=\frac{1}{cos \theta}=\frac{1}{\frac 1 3}=3 sec θ = cos θ 1 = 3 1 1 = 3
sin θ = − y r = − 8 3 \sin \theta =-\frac y r=- \frac {\sqrt{8}}3 sin θ = − r y = − 3 8
csc θ = 1 s i n θ = 1 − 8 3 = − 3 8 \csc \theta =\frac 1 {sin\theta}=\frac {1}{-\frac {\sqrt 8}{3}}=-\frac{3}{\sqrt 8} csc θ = s in θ 1 = − 3 8 1 = − 8 3
tan θ = − y x = − 8 1 = − 8 \tan \theta =-\frac y x=-\frac {\sqrt 8} 1=-\sqrt 8 tan θ = − x y = − 1 8 = − 8
cot θ = 1 t a n θ = − 1 8 \cot \theta=\frac{1}{tan \theta}=-\frac{1}{\sqrt 8} cot θ = t an θ 1 = − 8 1
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