Help me prove tanx/secx-1 = secx + 1/ tanx?

Answer 1

LHS
= tan x / (sec x - 1)
= tan x (sec x + 1) / (sec x - 1)(sec x + 1)
= tan x (sec x + 1) / (sec^2 (x) -1)
= tan x (sec x + 1) tan^2 (x)
= (sec x + 1) / tan x
=RHS

Answer 2

tanx/(secx-1) = (secx + 1)/ tanx <-- I think you mean this. If so,
sec^2 x - 1 = tan^2 x
sec^2x = 1 +tan^2x which is a well-known identity

If you wish further proof, then:
1/cos^2x = 1 + sin^2x/cos^2x
= (cos^2x+sin^2x)/cos^2x = 1/cos^2x

Answer 3

I can't see how what you've written is right? Maybe you made a mistake with the brackets?

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

It's very important you put brackets around everything unless it's very very clear what you mean.

Answer 4

tanx+secx-a million/tanx-secx+a million=(a million+sin... multiply the Nr and Dr via (tanx-secx)-a million [tanx-a million+secx)][(tanx-a million-secx)]/... =(tanx-a million)^2-sec^2x/(tanx-secx)... =a million+tan^2x-2tanx-sec^2x/tan^2x+... =-2tanx/2tan^2x-2tanxsecx =-a million/tanx-secx =-a million/(sinx/cosx)-(a million/cosx) =-1cosx/sinx-a million =cosx/a million-sinx multiplying theNr and Dr by1+sinx =cosx(a million+sinx)/(a million-sinx)(a million+sinx) =cosx(a million+sinx) a million-sin^2x =cosx(a million+sinx)/cos^2x =a million+sinx/cosx

Answer 5

tanx/(secx - 1)

(sinx/cosx)/(1/cosx - 1)

sinx/(1 - cosx) {*cosx/cosx}

[sinx(1 + cosx)]/[1 - cos^2] {*(1 + cosx)/(1 + cosx)}

[sinx(1 + cosx)]/sin^2x

(1 + cosx)/sinx

(1/cosx + 1)/(sinx/cosx)

(secx + 1)/tanx

I guess I took a long way around

Answer 6

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