Gary D. answered 11/11/15
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Gary, Math and Science Tutor Chicago, IL
Check the problem again!
Cosecant (csc) is the reciprocal of sine (sin) so they are always either both positive or both negative.
Perhaps it should be sine < 0 and cosine > 0?
In that case, sine is less than zero in quadrants 3 and 4, and cosine is greater than zero in quadrants 1 and 4, so this angle can only lie in quadrant 4.
On the unit circle, remember that cosine is the x-coordinate of the terminal side of the angle and sine is the y-coordinate. Quadrant 1 is that where both sine and cosine are greater than zero. The rest of them are numbered consecutively going counter-clockwise; so quadrant 2 has cos < 0 and sin > 0, quadrant 3 has cos < 0 and sin < 0, and quadrant 4 has cos > 0 and sin < 0