2007-03-13 19:50:23 · 7 answers · asked by QED 1 in Science & Mathematics ➔ Mathematics
The derivative of cos(x) is -sin(x).
sec(x)/(csc(x)*tan(x) = (1/cosx)/((1/sinx)*(sinx/cosx)) = (1/cosx)/(1/cosx) = 1.
Or if you meant [sec(x)/csc(x)] * tanx...
(1/cosx)/(1/sinx) * tanx = sinx/cosx * tanx = (tan^2)*x.
2007-03-13 19:56:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The derivative of cos x is -sin x.
sec x/csc x · tan x is the same as tan² x, which is not a derivative of any of the 6 basic circular functions.
^_^
2007-03-14 03:00:33 · answer #2 · answered by kevin! 5 · 0⤊ 0⤋
You are in big trouble here. The derivative of cos(x) = -sin(x);
sec(x)/csc(x) = sin(x)/cos(x) = tan(x). This time tan(x) = tan^2(x), not -sin(x); unless you meant sec(x)/[csc(x)*tan(x)] in which case the answer is 1, still not -sin(x).
2007-03-14 02:57:37 · answer #3 · answered by gp4rts 7 · 0⤊ 0⤋
the derivative of cos(x) is just -sin(x)
so you are in big trouble...........
sec(x)/csc(x)*tan(x) is equal to 1
2007-03-14 02:55:44 · answer #4 · answered by vatsa 2 · 0⤊ 0⤋
I hope that that simplifies to -sin(x), because that is the correct answer.
2007-03-14 02:54:05 · answer #5 · answered by Anonymous · 0⤊ 0⤋
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2007-03-14 02:52:08 · answer #6 · answered by GodsOfQED 1 · 0⤊ 2⤋
ummm it's -sin(x)
2007-03-14 02:52:59 · answer #7 · answered by Anonymous · 0⤊ 0⤋