Verified answer

tan theta = y/x = -1

in QII, x < 0, y > 0

x = -1, y = 1

hypotenuse = √(x^2 + y^2) = √(1 + 1) = √2

sinx = y/h = 1/√2 = √2/2

cosx = x/h = -1/√2 = -√2/2

cotx = x/y = -1/1 = -1

secx = h/x = -√2/1

cscx = h/y = √2/1

And here is explanation reply to your comment (tan theta = sqrt3, QIII)

tan theta = sqrt3, tan theta = y/x = √3

y = √3 x

In QIII, x < 0 and y < 0

x = -1, y = -√3, r = √[(-√3)^2 + (-1)^2] = 2

sin theta = y/r = -√3 / 2 = -√3/2

cos theta = x/r = -1/2

cot theta = x/y = -1/(-√3) = √3/3

"How do I draw"

Put one vertex at the origin,

one vertex in the 2nd quadrant along the major diagonal (y = -x),

and the third vertex on the x-axis directly below the 2nd vertex.

We needn't insist on a "unit circle," so just make the vertices

(0,0), (-1,0), and then because tan(theta) = -1, the upper vertex is (-1,1).

Pythagorean Theorem says r = sqrt(1 + 1) = sqrt(2).

Now use the definitions you mentioned in your question.

y = 1, x = - 1 and r = √2

You should be able to find the other five ratios

sin(θ) = 1/√2 = √2/2

cos(θ) = -1/√2 = -√2/2

csc(θ) = √2

sec(θ) = - √2

cot(θ) = - 1

tan θ = 1

Ignore the sign for a moment

In the second quadrant, "sin is positive, cos and tan are negative."

tan θ = -1

tan = opp/adj = 1/1 = 1

opp = 1

adj = 1

hyp = sqrt(1^2 + 1^2) = sqrt(2) (Pythogorean theorem)

sin θ = opp/hyp = 1/√2

cos θ = adj /hyp = -1/√2

tan θ = opp/adj = -1

coθ θ = 1/tan = -1

sec θ = 1/cos = -√2

csc θ = 1/sin = √2

https://gyazo.com/bdb23a8901a7f249cae81ce1e8b5ff04

Sides are 1 , 1 , √2 where √2 is the hypotenuse.

sin ∅ = 1/√2_________csc ∅ = √2

cos ∅ = - 1/√2_______sec ∅ = - √2

tan ∅ = - 1___________cot ∅ = - 1