Trigonometric Functions

Determine the Value of all six trigonometric functions of theta when given?
I need help. This is for my midterm review, if anyone is willing to help, I'd appreciate it.

a) sin theta= 3/5 and theta is in quadrant 1.
b) tan theta= -2 and theta is in quadrant II
c) sec theta = 2 radical 3/3 and theta is in quadrant II
d) csc theta= -2/3 and theta is in quadrant III.

I need to find all six trigonometric functions for each one ( sin, cos, tan, csc, sec, cot)

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  1. With each function given, you have two sides of a right triangle. Using the pyth theorm, you can find the other side, and thus, the other functions. We will be happy to check your work.

    For example, c) If secTheta = 2 sqrt (1/3), then the sides of the triangle are

    opposite figure out
    adjacent sqrt 3
    hypo= 2

    and the opposite will be

    4=3+opposite^2 or opposite is +-1, Now in quadrant two, that means sec is negative, so it should have been -2sqrt(1/3), and the opposite (vertical) is 1. From that you get the six functions.

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    bobpursley
  2. I will do c. You try the rest.
    Oh no I won't --> sign of sec is negative in quad 2. I think typo
    I will do d.
    d) csc T = 1/sin T = -2/3
    Hey! what is going on here?
    the absolute value of sin T may not be greater than 1. It can not be -3/2
    Either we are looking at typos or someone is playing games with you.
    I will do b I guess
    b) draw triangle in quad 2
    -1 along -x axis
    +2 up at x = -2
    now hypotenuse = sqrt (-1^2 + 2^2) = sqrt 5
    so
    sin = 2/sqrt 5
    cos = -1/sqrt 5 = -(1/5)sqrt 5
    tan = -2/1
    csc = 1/sin = (1/2)sqrt 5
    sec = 1/cos = -sqrt 5
    cot = 1/tan = -1/2

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  3. prove the identities (theta + phi), cos theta = -1/3 in Quadrant III, sin theta = 1/4 in quadrant II

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