If cot θ = −2, and θ terminates in QII, find the remaining trigonometric functions of θ.

Answer 1

cotx = -2 with x in the 2nd quadrant, which means sine and cosecant are >0, cosine and secant <0, tangent <0

cotx = adjacent side over opposite side = -1/2

tanx = 1/cotx = 2/-1 = - 2

hypotenuse squared = sum of squares of adjacent & opposite sides

h^2 = 2^2 + 1^2 = 5

h = sqr5

sinx = opposite side over hypotenuse = 2/sqr5

cscx = 1/sinx = sqr5/2

cosx = adjacent side over hypotenuse = -1/sqr5

secx = 1/cosx = -sqr5