Placement question #14
assessment progress
42 percent
let
θ
be an angle in quadrant
iv
such that
=cosθ45
.
find the exact values of
cscθ
and
cotθ

Your statement is not clear.
still if ∅=45
and ∅ lies in 4th quadrant.
csc&cot  are negative in 4th quadrant.
csc ∅=1/sin ∅=1/+-sqrt(1-cos^2 ∅)
=-1/sqrt(1-cos^245)
=-1/sqrt(1-(1/√2)^2)
=-1/√(1-1/4)
=-1/√(3/4)
=-2/√3
cot ∅=cos ∅/sin∅=cos315/sin315=cos(360-45)/sin (360-45)
=cos 45/-sin 45=-1

15-7=8         
i hope it helps