more ALG2/TRIG HELP!!?

Question

(cos x) - (cos x / 1 - tan x) = (sin x cos x) / (sin x - cos x)

i keep getting stuck at either cos x - sin x -sinx or 0.

Answer 1

hi again,

i hope u have tried a liitle to solve your own problem.

(cos x) - (cos x / 1 - tan x) = cos x (1 - tan x) / (1 - tan x) - (cos x / 1 - tan x)

note: 1- tan x = (cos x - sin x) / cos x
and cos x (1 - tan x) = cos x - sin x

so

cos x (1 - tan x) / (1 - tan x) - (cos x / 1 - tan x)
= (cos x - sin x ) / (1 - tan x) - (cos x / 1 - tan x)
= - sin x / (1 - tan x)
= - sin x / [(cos x - sin x) / cos x]
= -sin x cos x / (cos x - sin x )
= sin x cos x / (sin x - cos x)

Answer 2

cos x - sin x= 1

Answer 3

cos x - (cos x/(1- tan x))
take cos x common

= cos x ( 1- 1/(1-tan x))

convert tan x to sin x/ cos x as no tn x on RHS
= cos x(1-1/(1- sin x/cos x))
simplify
= cos x (1- cos x/(cos x- sin x))
convert cos x - sin x to sin x - cos x as on RHS
= cos x ( 1 + cos x/( sin x- cos x)))
= cos x(sinx - cos x + cos x)/(sin x- cos x)
= cos x. sin x/(sin x- cos x)
=RHS
QED

Answer 4

LHS=
tanx=sinx/cosx put this instead of tanx
u will get
cosx-(cosx/1-(sinx/cosx))
cosx-cosx/((cosx-sinx)/cosx)
take cosx in denominator to numerator
cosx-(cosx)^2/(cosx-sinx)
(cosx^2-cosxsinx-cosx^2)/(cosx-sinx)
-cosxsinx/-(sinx-cosx) taking -ive sign common
-ive sign cancel out in the above expression
u get
sinxcosx/(sinx-cosx)
hence
LHS=RHS

Answer 5

(cosx)- (cosx/(1-tanx)
(cosx)- (cosx)^2/(cosx -sinx)
((cosx)(cosx-sinx)-(cosx)^2)/(cosx-sinx)
((cosx)^2-cosxsinx-(cosx)^2)/(cosx-sinx)
-cosxsinx/(cosx-sinx)
cosxsinx/(sinx-cosx)
sinxcosx/(sinx-cosx)

the part after the ... is just (cosx-sinx)