Any would be greatly appreciated. i am completely clueless on all of this stuff. we recently got a new teacher and she doesn't know what she's doing, and we have a huge test tomorrow on this. these are some of the study guide questions that i am completely clueless on.

1.) determine the quadrant in which θ lies: tanθ< 0, cosθ> 0
2.) given sinθ= -1/5 and tan θ< 0, find cosθ
3.) find the reference angle for θ=228°
4.) use the reference angle to find the exact value of cot(-150°)
5.) use the reference angle to find the exact value of csc(225°)
6.) use the reference angle to find the exact value of sin(- 8π/3)
7.) use the reference angle to find the exact value of sec(7π/4)
8.) given cosθ= -7/8 and π/2 ≤ θ ≤ π, find tanθ
9.) given sinθ= -4/7 and 3π/2 ≤ θ ≤ 2π, find secθ

When a triangle is formed by a ray starting at the origin and making some angle with the x-axis, we let y-component be the opposite to the angle or the rise of the triangle and let the x-component be the adjacent of the angle or the run. Hypotenuse (r) is always positive.
QUESTION 1:
So for cos Ф ≥ 0 (sorry for the symbol) you have cos Ф=adjacent/hypotenuse so CosФ=x/r oviously x have to be positive because r is positive all the time and you need  equal symbols for ≥ 0.
And for tanФ≤0 we know tanФ=opposite/adjacent i mean tanФ=y/x you kno now that x is positive from above then y must be negative because for getting ≤0 the symbols have to be differents, so we have x positive and y negative we are in Quadrant IV.
QUESTION 2:
From Pythagorean we know r^2=adj^2+opp^2
we have sinФ=-1/5, sinФ=opp/r... so opp=y=-1 and r=5
adj^2=r^2-opp^2  adj=root(5^2-(-1)^2)=root(24)=2root(6)=x
Now if tanФ≤0 we have tanФ=opp/adj=y/x. From senФ=-1/5 we know y=-1 so x must be positive because ≤0
Solve cosФ=adj/r=(2root(6))/5=0.98
QUESTION 3:
reference angle for Θ=228°
the reference angle It is the angle formed with the x axis in the quadrant where the line formed by the angle is given, in this case the angle is in quadrant III so you have to substract 228-180=48
QUESTION 4:
Draw the angle from positive x-axis down so for the reference add -150+180=30, from trig table we know cos(30)=root(3)/2 but the angle -150 is in quadrant III so cosФ=x/r then x must be negative so cos(-150)=-root(3)/2.
I hope that I can continue tomorrowif you want. Sorry is too late for me now.