2006-07-27 11:29:51 · 10 answers · asked by Ron DMC 2 in Science & Mathematics ➔ Mathematics
cos x (1/sinx) = cot x
cos x / sin x = cot x
cot x = cot x
2006-07-27 11:34:43 · answer #1 · answered by jaredkomahalko 2 · 0⤊ 3⤋
Sometimes it helps to state everything in terms of sin & cos
cos x csc x = cot x
But csc x = 1/sinx and cotx = cosx/sinx
So
cos x/sin x = cos x/sin x
2006-07-27 18:37:00 · answer #2 · answered by kindricko 7 · 0⤊ 0⤋
cos x * csc x = cot x
cos x * (1/sin x) = cot x
cos x / sinx = cot x.
That is the definiton of cot x, so that proves the equation.
2006-07-27 20:37:50 · answer #3 · answered by Anonymous · 0⤊ 0⤋
csc x = 1/sin x
cot x = cos x/sin x
cos x csc x = cos x/ sin x
Thus, = cot x.
2006-07-27 18:34:47 · answer #4 · answered by kooshman38 3 · 0⤊ 0⤋
csc x = 1/(sin x)...by definition
(cos x)(1/sinx)=(cos x)/(sin x)...by multiplication
(cos x)/(sin x)=1/(tan x)...by definition
1/(tan x)=cot x...by definition
therefore, (cos x)(csc x)=cot x...by substitiution
2006-07-27 18:36:33 · answer #5 · answered by jogimo2 3 · 0⤊ 0⤋
Change to sines and cosines, then change back... it's not too bad. :-)
cos(x) · csc(x)
= cos(x) · [1 / sin(x)]
= cos(x) / sin(x)
= cot(x)
2006-07-27 18:33:55 · answer #6 · answered by Louise 5 · 0⤊ 0⤋
(cos x)(csc x) = (cot x)
(cos x)(1/sin x) = (cos x/sin x)
(cos x/sin x) = (cos x/sin x)
2006-07-27 18:36:10 · answer #7 · answered by Anonymous · 0⤊ 0⤋
cosx = a/h
cscx = h/a
cotx = a/o
=> a/h * h/o = ah/ho = a/o
where a:= adjacent
h:= hypotenuse
o:= opposite
2006-07-27 18:42:58 · answer #8 · answered by UROQ 2 · 0⤊ 0⤋
cosxcscx = cotx
cscx = (1/sinx)
(cosx)(1/sinx) = cotx
(cosx/sinx) = cotx
cotx = cotx
2006-07-27 19:44:03 · answer #9 · answered by Sherman81 6 · 0⤊ 0⤋
change all of them to sines and cosines:
cosx(1/sinx)=cosx/sinx
cosx/sinx=cosx/sinx
2006-07-27 18:59:41 · answer #10 · answered by pilotmanitalia 5 · 0⤊ 0⤋