determine the values of the other trigonometric functions Solution ----------------------- In pi/2 < x < pi, sinx will be -ve. sin^2(x) = 1 - cos^2(x) = 1 - (-2/3)^2 = 5/9 sinx = -sqrt(5) /3 --- tan(x) = sinx / cosx = sqrt(5) / 2 --- csc(x) = 1/sinx = -3 / sqrt(5) --- sec(x) = 1/cosx = -3/2 --- cot(x) = cosx / sinx = 2/sqrt(5).