(cot x) / (csc (-x))?

Question

(cot x) / (csc (-x))
= (1/tan x) / ( 1/sin -x)
=(1/tan x) X (sin -x / 1) ????
?
The answers given are
tan x
sin 2x
-cos x
cos x
or -cos^2x

any ideas on my next step?

Aha! forgot to make 1/tan x = cos x/ sin x thanks for the help! you all rock!

Answer 1

(cot x) / (csc (-x))
= (cosx/sinx)(-sinx)
= -cosx

Answer 2

I convert problems like these to sines and cosines. So:
cot(x)/csc(-x) = [cos(x)/sin(x)] / [1/sin(-x)]

Given that sin(-x) = -sin(x), you should be able to finish from here.

In circle-based trigonometry, this is automatic because only cos(x) and sin(x) are DEFINED, and the other functions are DERIVED:
tan(x) = sin(x)/cos(x)
sec(x) = 1/cos(x)
csc(x) = 1/sin(x)
cot(x) = cos(x)/sin(x)

If you are in an introductory class using a triangle-based approach, it is very helpful to memorize those four equations as Important Identities.

Answer 3

(cot x) / (csc (-x))
= (1/tan x) / ( 1/sin (-x))
= (1/tan x) X (sin (-x))
= (cos x/sinx)x(-sinx)
= -cosx

Even though i got beat to it..

Answer 4

(cot x) / (csc (-x))
= (1/tan x) / ( 1/sin (-x))
=(1/tan x) X (sin (-x))
=(cos x/sinx)x(-sinx)
= -cosx

Answer 5

cotx/csc-x
=(1/tanx)/(1/sin-x)
=(1/(sinx/cosx))/(1/-sinx)
=(cosx/sinx)*(-sinx)
= (-cosxsinx)/(sinx)
= -cosx