tan^2x/secx+1=1-cosx/cosx
i thought i got it to 1/cosx-1=1-cosx/cosx but im not sure if thats correct and dont know the next step.
2007-12-07 03:30:24 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
LHS
tan ² x / (sec x + 1)
(sec ² x - 1) 1 (sec x + 1)
(sec x - 1) (sec x + 1) / (sec x + 1)
sec x - 1
RHS
(1 - cos x) / cos x
(1/cos x) - 1
sec x - 1
LHS = RHS
2007-12-07 04:10:39 · answer #1 · answered by Como 7 · 4⤊ 0⤋
tan^2x/(secx+1)= (1-cosx)/cosx, Use parenthesis to make it clear.
tan^2x/(secx+1)
= (sec^2 x - 1)/(secx+1)
= secx - 1
= (1-cosx)/cosx
2007-12-07 03:37:25 · answer #2 · answered by sahsjing 7 · 2⤊ 0⤋
tan^2x/(secx + 1) = (sin^2x/cos^2x)/(secx + 1)
= sin^2x/ cos^2x(secX + 1)
= sin^2x / cosx + cos^2x
= (1-cos^2x) / (cosx + cos^2x)
factoring
=(1+cosx)(1-cosx) / cosx(1+ cosx)
= (1-cosx)/cosx
2007-12-07 03:40:56 · answer #3 · answered by Linda K 5 · 0⤊ 0⤋