Here’s my solution to a nice little trigonometric problem posted by miss loi.

Show that

[tex]\frac{ \tan x + \sec x – 1} { \tan x – \sec x +1 } \equiv \tan x + \sec x[/tex]

[tex]\frac{ \tan x + \sec x – 1} { \tan x – \sec x +1 } = \frac{ \sin x + 1 – \cos x } { \sin x – 1 + \cos x }[/tex]

[tex]= \frac{ \sin x + 1 – \cos x } { \sin x – 1 + \cos x } \times \frac{ \sin x + 1 + \cos x } { \sin x + \cos x + 1 } = \frac{ (\sin x + 1)^2 – \cos^2 x } { (\sin x + \cos x)^2 – 1 }[/tex]

[tex]= \frac{ \sin^2 x + 2 \sin x +1 – \cos^2 x } { 2 \sin x \cos x } = \frac{ 2 \sin^2 x + 2 \sin x } { 2 \sin x \cos x }[/tex]

[tex]= \tan x + \sec x[/tex]

(QED)

How to get an A1 if you never pass up your homework to Miss Loi?!

FYI, mimetex is also available there (she thinks!)for those who knows how to use it – just that sometimes those complex LaTex codes can be a little too profound for her bimbatic brain.

How about:

For [tex]\tan x – \sec x +1 \ne 0[/tex],

[tex](\tan x + \sec x – 1)/(\tan x – \sec x 1) = \tan x + \sec x[/tex]

if and only if

[tex]\tan x + \sec x – 1 = (\tan x + \sec x)(\tan x – \sec x + 1)[/tex]

[tex]= (\tan^2 x – \sec^2 x) (\tan x + \sec x)[/tex]

[tex]= \tan x + \sec x – 1[/tex]

Dear nick,

I took the liberty to edit your post to make use of the latexrender plugin. Yes, your proof is nice as well, although as a personal preference, I would avoid writing statements like A = A, so I would remove the

[tex] \tan x + \sec x – 1 [/tex]

in the first line.