Verify that: [cos(x)/1-sin(x)] - tan(x) = sec(x)?

Answer 1

[cos x / (1 - sin x)] - tan x
= cos x / (1 - sin x) - sin x / cos x
= cos^2 x - (sin x) (1 - sin x) / cos x (1 - sin x)
= (cos^2 x + sin^2 x - sin x) / cos x (1 -sin x)
= (1 - sin x) / cos x (1 - sin x)
= 1 / cos x
= sec x
Proved

Answer 2

LHS = cos x /(1- sin x) - tan x
= cos x(1+sin x)/(1-sin ^2 x) - tan x
= cos x(1+sin x)/cos^2 x -tan x
= (1+sin x)/cos x - sin x/cos x
= (1+sin x - sin x)/ cos x
= 1/ cos x
= sec x proved'

Answer 3

LHS = cos x / (1 - sin x) - tan x
= cos x / (1 - sin x) - sin x / cos x
= (cos x . cos x - sin x (1 - sin x)) / [cos x (1 - sin x)]
= (cos^2 x - sin x + sin^2 x) / [cos x (1 - sin x)]
= (1 - sin x) / [cos x (1 - sin x)]
= 1 / cos x
= sec x
= RHS.

Answer 4

cos/(a million-sin) = a million/cos + sin/cos cos/(a million-sin) = (a million + sin)/cos multiply the two factors with the aid of a million/(a million+sin): cos/(a million - sin^2) = a million/cos as a results of fact sin^2 + cos^2 = a million, cos^2 = a million - sin^2 : cos/cos^2 = a million/cos a million/cos = a million/cos wish that facilitates =)

Answer 5

LHS: (cos(x) * (1+sin(x))/[(1+sin(x))(1-sin(x))] - tan(x) =
(cos(x) * (1+sin(x))/(1-sin^2x) - tan(x) =
(cos(x) * (1+sin(x))/(cos^2x) - tan(x) =
(sec^2x)((cos(x) * (1+sin(x)) - sin(x)*sec(x) =
sec(x) * (cos(x)sec(x) + cos(x)sec(x)sin(x) - sin(x)) =
sec(x) *(1 + sin(x) - sin(x)) =
sec(x) = RHS

Answer 6

change LHS to cosine and sine only, then find LCD to add together:

cosx/(1-sinx) - sinx/cosx
cos^2(x) / (cosx(1-sinx)) - sinx(1-sinx) / (cosx(1-sinx))
[ cos^2x - sinx(1-sinx) ] / [cosx(1-sinx) ]
[cos^2x-sinx+sin^2x ] / [ cosx(1-sinx) ] since cos^2+sin^2 =1

(1-sinx) / [cosx(1-sinx)]
cancel same term
1/cosx
which is secx

:)