Question

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Answer 1

I had a problem very similar to this that i was working today. lol
remember that csc=1/sin
cot= y/x or sin/cos
\[\sin ^{2}\]=csc

Answer 2

you should be able to go from there. Just remember your trig identities its key to solving problems such as these! I hopes this helps. also remember:
cos=x/r
sin=y/r
tan=slope=y/x
sec=1/cos
csc=1/sin
cot=x/y
sometimes writing them at the top of your paper helps when you feel stressed. :)

Answer 3

@johnny101

Are you familiar with these identities? See attachment.

Answer 4

Take a look at the third one.

1 + cot² x = csc² x where x = theta

Given: csc (theta) = 3/2

1 + cot² x = (3/2)²

What comes next? @johnny101

Answer 5

csc= 3/2 (this was given to you)
csc=1/sin
sin theta= 2/3
sin theta= y/r
so use x^2+y^2=r^2 to find the value of x
make sure to draw the triangle it helps. draw from the origin out into whatever quadrant it is in and label the side parallel to the y-axis as y=2, label the hypotenuse= r= 3, and label the x-axis as x=sq rt 5
now you have every thing that you need
cot=x/y= sqrt 5/2

Answer 6

you do not have to use the identities for this problem since it appears that you have not learned the double angle identities yet. make sure that you still remember them though and when in doubt draw a triangle label the sides and use x^2+y^2=r^2 it may take a little longer but it wont steer you wrong.