Determine the quadrant when the terminal side of the angle lies according to the following conditions: sin (t) > 0, cos (t) < 0.

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

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4