another one TRIG.?

Question

( secx - tanx)^2 = (1-sinx)/ ( 1+sinx )

Answer 1

S.T.( secx - tanx)^2 = (1-sinx)/ ( 1+sinx )
to prove relation we must start by either left hand side or right hand side :
L.H.S=( secx - tanx)^2
=(1/cosx-sinx/cosx)^2 combining denominators
=( (1-sinx) / cos x )^2 raising to power 2
= (1-sinx )(1-sinx ) / cos^2x
cos^2x = 1- sin^2x ( sub)
L.H.S = (1-sinx )(1-sinx ) / (1-sin^2x)
= (1-sinx )(1-sinx ) / (1-sinx ) (1+sinx )
deleting (1-sinx ) up and down
L.H.S. (1-sinx ) / (1+sinx ) which is exactly R.H.S.
HOPE THAT HELPS

Answer 2

I think your question is like this
S.T.( secx - tanx)^2 = (1-sinx)/ ( 1+sinx )
Ans:
LHS=( secx - tanx)^2
=[1/Cosx-Sinx/Cosx]^2
=[(1-Sinx)^2/Cosx^2]
=[(1-Sinx)^2/(1-Sinx)(1+SinX)]
By cancelling (1-Sinx) we get
=(1-Sinx)/(1+Sinx)
Good Luck

Answer 3

( secx - tanx)^2 = (1-sinx)/ ( 1+sinx )
sec^2(x)-2(Sinx/cos^2(x))+tan^2x=(1-sin^2(x)^2/cos^2(x)
1-2sinx+sin^2(x)=(1-sinx)^2
no solutions

Answer 4

remember secx=1/cosx, and tanx=sinx/cosx
(1/cosx-sinx/cosx)^2
([1-sinx]/cosx)^2 combine fraction
[1-sinx]^2/ cos^2x
[1-sinx]^2/(1-sin^2x) cuz sin^2x+cos^2x=1 so cos^2x=1-sin^2x
[1-sinx]^2/[1-sinx][1+sinx] cuz a^2-b^2=(a-b)(a+b)
cancel 1-sinx
answer is [1-sinx]/[1+sinx] prove it that is equal right hand side.
Note cos^2x is cosine square of x not cos2x, also with sin^2x that is sine square of x.
good luck

Answer 5

(1/cosx - sinx/cosx)^2 = (1 - sinx)/(1 + sinx)
((1 - sinx)/cosx)^2 = "
((1 - sinx)(1 - sinx))/cos^2 x = "
((1 - sinx)(1 - sinx))/(1 - sin^2 x) = "
((1 - sinx)(1 - sinx))/((1 + sinx)(1 - sinx)) = "
(1 - sinx)/(1 + sinx) = (1 - sinx)/(1 + sinx).