2007-11-18 10:51:35 · 8 answers · asked by haru 2 in Science & Mathematics ➔ Mathematics
2(1 - sin ² x) / cos x
2 cos ² x / cos x
2 cos x
2007-11-22 06:55:45 · answer #1 · answered by Como 7 · 3⤊ 1⤋
2
2007-11-18 18:54:59 · answer #2 · answered by Anonymous · 0⤊ 1⤋
factor a 2 from the numerator and use the identity cos^2(x) + sin^2 (x) = 1
final answer should be 2cosx
2007-11-18 18:54:37 · answer #3 · answered by Anonymous · 1⤊ 0⤋
(2 - 2sin^2 x) / cos x =
2(1 - sin^2 x) / cos x =
2cos^2 x / cos x =
2cos x
2007-11-18 18:55:00 · answer #4 · answered by UnknownD 6 · 1⤊ 1⤋
expression 2-2sinx^2/cosx
you can pull out a 2 2(1-sinx^2)/cosx
1-sinx^2=cosx^2 so, 2(cosx^2)/ cosx
2(cosx)(cosx)/cosx cross out a set of cosx
Answer = 2cosx
2007-11-18 19:01:48 · answer #5 · answered by makesmegiggle011 3 · 1⤊ 0⤋
2-2sin^2x over cosx =
2-2(1-cos^2x) over cosx= (because sin^2x+cos^2x =1)
2-2+2cos^2x over cosx=
2cos^2x over cosx=
2cosx
:]
2007-11-18 18:56:28 · answer #6 · answered by My Lovee 3 · 1⤊ 1⤋
for every angle x , there is sin²x+cos²x=1
so sin²x=1-cos²x
so
(2-2sin²x)/cosx=
[2-2(1-cos²x)]/cosx=
=(2-2+2cos²x)/cosx=
=2cos²x/cosx=
=2cosx
2007-11-18 19:02:10 · answer #7 · answered by Kulubaki 3 · 1⤊ 1⤋
[2 -- 2sin^2x ]/cosx
=2(1 --sin^2x)/cosx
=2cos^2x/cosx
= 2cosx
2007-11-18 18:56:18 · answer #8 · answered by sv 7 · 2⤊ 1⤋