2006-12-15 02:43:26 · 5 answers · asked by marshie 1 in Science & Mathematics ➔ Mathematics
NUmerator = tan x + sec x -1
= sin x/cos x + 1/cos x - 1
= (sin x + 1)/cos x -1
denominator = (sin x -1)/cos x +1
multiply both by cos x
numerator = sin x +1 - cos x
= 2 sin x/2 cos x/2 + 2 sin^ x/2
denominator = 2 sin x/2 cos x/2 - 2 sin ^2 x/2
devide
number = (cos x/2 + sin x/2)/(cos x/2 - sin x/2)
= (cos x/2 + sin x/2)^2/(cos^2 x/2- sin ^2 x/2)
= (cos^2x/2+ sin^2 x/2 + 2 cos x/2 sin x/2)/(cos x)
= (1+ sin x)/ cos x
2006-12-18 01:15:01 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
tanx+secx-1/tanx-secx+1=(1+sin...
multiply the Nr and Dr by (tanx-secx)-1
[tanx-1+secx)][(tanx-1-secx)]/...
=(tanx-1)^2-sec^2x/(tanx-secx)...
=1+tan^2x-2tanx-sec^2x/tan^2x+...
=-2tanx/2tan^2x-2tanxsecx
=-1/tanx-secx
=-1/(sinx/cosx)-(1/cosx)
=-1cosx/sinx-1
=cosx/1-sinx
multiplying theNr and Dr by1+sinx
=cosx(1+sinx)/(1-sinx)(1+sinx)
=cosx(1+sinx)
1-sin^2x
=cosx(1+sinx)/cos^2x
=1+sinx/cosx
2006-12-17 04:31:22 · answer #2 · answered by arpita 5 · 0⤊ 0⤋
tanx+secx-1/tanx-secx+1=(1+sinx)/cosx
multiply the Nr and Dr by (tanx-secx)-1
[tanx-1+secx)][(tanx-1-secx)]/(tanx-secx+1)(tanx-secx-1)
=(tanx-1)^2-sec^2x/(tanx-secx)^2-1
=1+tan^2x-2tanx-sec^2x/tan^2x+sec^2x-2tanxsecx-1
=-2tanx/2tan^2x-2tanxsecx
=-1/tanx-secx
=-1/(sinx/cosx)-(1/cosx)
=-1cosx/sinx-1
=cosx/1-sinx
multiplying theNr and Dr by1+sinx
=cosx(1+sinx)/(1-sinx)(1+sinx)
=cosx(1+sinx)
1-sin^2x
=cosx(1+sinx)/cos^2x
=1+sinx/cosx
2006-12-15 10:57:32 · answer #3 · answered by raj 7 · 0⤊ 0⤋
It appears the end of your problem got cut off. You probably need to put some spaces in there so that doesn't happen.
2006-12-15 10:52:15 · answer #4 · answered by Jim Burnell 6 · 0⤊ 0⤋
=TAN2X-SEC2X-1/TANX-SECX+1
=0+1
=1
2006-12-16 03:58:49 · answer #5 · answered by Anonymous · 0⤊ 0⤋