suspect you might mean
sec x / (sec x -tan x) = sec^2 x + sec x + tan x
work on the left
1/cos x / [ 1/cos x -sin x/cos x ] =
1/[1 - sin x] =
[ 1 + sin x ]/ [1-sin^2 x] =
[1+sin x ]/cos^2 x =
work on the right
= 1/cos^2 x + 1/cos x + sin x/cos x)
= (1/cos^2 x) [1 + cos x + sin x cos x]
sorry, I can not make that come out. Please check for typos
Sorry, it was a typo, it's supposed to be:
LS = (1/cosx)/(1/cosx - sinx/cosx)
= (1/cosx)/( (1-sinx)/cox )
RS = 1/cos^2 x + (1/cosx)(sinx/cosx)
= 1/cos^2 x + sinx/cos^2 x
= (1+ sinx)/cos^2x
back to LS
LS = 1/(1-sinx) * (1+sinx)/(1+sinx)
= (1+sinx)(1 - sin^2 x)
= (1+sinx)/cos^2 x
Well then we are finished
work on the right
= 1/cos^2 x + (1/cos x * sin x/cos x)
= (1/cos^2 x) [1 + sin x]
:)thanks!!! I've been trying to figure this out for a while now, can you explain this to me: what do I do when i get to something like this
If you see (sin x - 1) on the bottom try multiplying top and bottom by (sin x + 1)
1/(sin x - 1) * [ (sin x + 1)/(sin x+1) ]
= (sin x + 1) / (sin^2 x - 1)
= (sin x +1)/-cos^2 x
Answer this Question
precal - 1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-...
Math - I can't find the integral for (tanx)^(6)*(secx)^(2) I tried splitting up ...
Calculus - find the derivative of f(x)=tanx-4/secx I used the quotient rule and ...
Calculus - Int tanx sec^2x dx can be taken as (by putting it in form of Int xdx...
math - 1. (sinx/cscx)+(cosx/secx)=1 2. (1/sinxcosx)-(cosx/sinx)=tanx 3. (1/1+cos...
calculus - find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx ...
Calculus - Integration - Hello! I really don't think I am understanding my calc ...
precal - (1/tanx-secx))+(1/(tanx+secx))=-2tanx
Trig - Verify the identity. (secx + tanx)/(secx - tanx) = (1 + 2sinx + sin(^2)x...
Math - Im really struggling with these proving identities problems can somebody ...