2007-04-23 10:02:47 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(sin x/cos x) / (sin x cos x)
= (sin x / cos x) * (1/ sin x cos x)
= (sin x) / (sin x cos x cos x)
= 1 / (cos x)^2
= (1 / cos x)^2
= (sec x)^2
2007-04-23 10:07:14 · answer #1 · answered by Mathematica 7 · 0⤊ 1⤋
there is two residences utilized right here, which i assume you already be attentive to: a million. (a+b)(a-b) = a^2-b^2 2. cos(2x) = cos^2x-sin^2x So lower back on your equation, utilising the 1st assets: (sinx-cosx)(sinx+cosx) = sin^2x-cos^2x Your equation now turns right into a million+sin^2x-cos^2x, or a million-(cos^2x-sin^2x). Now practice the 2nd assets, changing the section in parantheses, and you get: a million-cos2x.
2016-12-16 13:40:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
this is what I got, I'm not sure if it's what you're looking for:
(sinx/cosx)/(sinxcosx) multiply top and bottom by cosx
(sinx)/(sinxcos^2x) then the sin's cancel
1/cos^2x = sec^2x
2007-04-23 10:08:10 · answer #3 · answered by rubiks87 2 · 0⤊ 1⤋
(sinx/cosx)/(sinx*cosx)= sinx/[sinx*(cosx)^2]= 1/(cosx)^2 = (secx)^2
2007-04-23 10:09:41 · answer #4 · answered by physicist 4 · 1⤊ 0⤋
=sinx/sinx.cos^2(x)
=1/cos^2(x)
=sec^2(x)
2007-04-23 10:06:56 · answer #5 · answered by Anonymous · 1⤊ 0⤋
sin x / cos x * 1 / (sin x cos x)
= 1 / cos^2x = sec^2 x
2007-04-23 10:06:05 · answer #6 · answered by Mαtt 6 · 1⤊ 1⤋
= (sin x / cos x ) / (sin x.cos x)
= (sin x) / (sin x.cos²x)
= 1 / cos²x
= sec²x
2007-04-23 10:08:53 · answer #7 · answered by Como 7 · 0⤊ 1⤋