cscx/cot^2x=(tanx)(secx)
How do I prove that trig identity while working with ONLY one side. I can't work both sides untl they agree.
2007-10-10 16:01:25 · 5 answers · asked by Keri 2 in Science & Mathematics ➔ Mathematics
csc(x) = 1/sin(x)
cot(x) = cos^(x)/sin^2(x)
(1 / sin(x)) / (cos^2(x) / sin^2(x))
sin^2(x) / sin(x) cos^2(x)
sin(x) / cos^2(x)
tan(x) sec(x)
Therefore csc(x) / cot^2(x) = tan(x) sec(x)
2007-10-10 16:08:00 · answer #1 · answered by Jeffrey R 3 · 0⤊ 2⤋
Note: CSCX= 1/sinx
COT^2 X= COS^2 X/ Sin^2 X
Therefore; 1/SINX/COS^2 X/ SIN^2 X =
(1/SINX)(SIN^2 X/COS^2 X) =
SINX/ COS^2X = (SINX/COSX)(1/COSX) =
SInce: SINX/COS = TANX and 1/COSX= SECX
therefore; (SINX/COSX)(1/COSX) = (TANX)(SECX)
2007-10-10 23:18:42 · answer #2 · answered by Pedro Q 3 · 0⤊ 1⤋
It's simple
csc x = 1/sinx
cot^2x = cos^2x / sin^2x
1/cot^2x = sin^2x / cos^2x
(sinx/cosx)^2 (1/sinx) = sinx / cos^2x
sin x / cos x = tan x
1/cos x = sec x
(sin x / cos x)(1/cos x) = sin x / cos^2x
(tan x) (sec x) = csc x / cot^2 x
2007-10-10 23:06:59 · answer #3 · answered by UnknownD 6 · 0⤊ 2⤋
cscx = 1/sinx
cot^2x = cos^2x / sin^2x
cscx / cot^2 x = (1/sinx) / (cos^2x / sin^2x)
= (1/sinx) * (sin^2x / cos^2x)
= (sinx) / (cos^2x)
= (sinx / cos x) * (1/cosx)
= (tanx) (secx)
2007-10-10 23:06:37 · answer #4 · answered by Anonymous · 1⤊ 2⤋
csc x/cot² x
csc x * tan² x
csc x * sin x/cos x * tan x
1/sin x * sin x/cos x * tan x
1/cos x * tan x
sec x tan x
2007-10-10 23:05:05 · answer #5 · answered by Pascal 7 · 0⤊ 2⤋