2006-09-25 08:07:52 · 7 answers · asked by Byron N 2 in Science & Mathematics ➔ Mathematics
sin^2(x)sec^2(x)-sin^2(x)?
2006-09-25 08:08:21 · update #1
I know the answer but I need someone to explain it
2006-09-25 08:13:57 · update #2
Factor out the sine function:
sin^2 x * sec^2 x - sin^2 x =
... = sin^2 x * (sec^2 x - 1)
... = sin^2 x * (1/cos^2 x - 1)
... = sin^2 x * (1 - cos^2 x) / cos^2 x
... = sin^2 x * sin^2 x / cos^2 x
... = sin^4 x / cos^2 x
OR
... = sin^4 x / (1 - sin^2 x)
2006-09-25 08:14:09 · answer #1 · answered by dutch_prof 4 · 0⤊ 0⤋
We start with:
sin² x sec²x - sin² x
We can factor out the sin² x:
sin² x ( sec² x - 1 )
Since tan² x + 1 = sec² x, we get:
sin² x tan² x
Since tan x = sin x / cos x, we could also write this as:
sin^4 x / cos² x
Hope that helps!
2006-09-25 08:15:04 · answer #2 · answered by Jay H 5 · 0⤊ 0⤋
sin^2 x* sec^2 x-sin^2 x
=sin^2 x(sec^2 x-1)
=sin^2 x(tan^2 x+1-1)
=sin^2 x*tan^2 x
2006-09-25 08:16:21 · answer #3 · answered by openpsychy 6 · 0⤊ 0⤋
factor out the sin^2(x) to get:
sin^2(x)[sec^2(x) - 1]
which I believe is:
sin^2(x)tan^2(x) = sin^4(x)/cos^2(x)
2006-09-25 08:15:05 · answer #4 · answered by ohmneo 3 · 0⤊ 0⤋
you know that sec^2(x)=1/cos^2(x) so the equation became tan^2(x)-sin^2(x)
2006-09-25 08:24:28 · answer #5 · answered by Anonymous · 0⤊ 0⤋
sin^2x(sec^2x-1)
=sin^2x*tan^2x
2006-09-25 08:15:02 · answer #6 · answered by raj 7 · 0⤊ 0⤋
ummmmm. NO. :)
2006-09-25 08:11:22 · answer #7 · answered by beckray 4 · 0⤊ 1⤋