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Topic: Proving trigonometric identities
Replies: 32   Last Post: Jan 27, 2009 1:59 AM

 Messages: [ Previous | Next ]
 Dave Dodson Posts: 690 Registered: 12/13/04
Re: Proving trigonometric identities
Posted: Jan 23, 2009 10:10 AM

On Jan 23, 2:41 am, Albert <albert.xtheunkno...@gmail.com> wrote:
> This (I think) is the last trig proof I've to do (there may be some in
> the next booklet though):
>
> (tan x + sec x - 1) / (tan x - sec x + 1) = (1 + sin x) / cos x

The secret to this is to multiply numerator and denominator by
something that creates a Pythagorean identity in the denominator:

(tan x + sec x - 1) / (tan x - sec x + 1)
= (tan x + sec x - 1) / (tan x - sec x + 1) * (tan x + sec x + 1) /
(tan x + sec x + 1)
= [(tan x + sec x) - 1] * [(tan x + sec x) + 1] / {[(tan x + 1) - sec
x] * [(tan x + 1) + sec x]}
= [(tan x + sec x)^2 - 1] / [(tan x + 1)^2 - sec^2 x]
= [tan^2 x + 2 * tan x * sec x + sec^2 x - 1] / [tan^2 x + 2 * tan x +
1 - sec^2 x]
= [tan^2 x + 2 * tan x * sec x + tan^2 x] / [tan^2 x + 2 * tan x -
tan^2 x]
= [2 * tan^2 x + 2 * tan x * sec x] / [2 * tan x]
= tan x + sec x
= sin x / cos x + 1 / cos x
= (1 + sin x) / cos x

Dave

Date Subject Author
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Angus Rodgers
1/20/09 Guest
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 matt271829-news@yahoo.co.uk
1/21/09 Albert
1/21/09 matt271829-news@yahoo.co.uk
1/23/09 Albert
1/23/09 Angus Rodgers
1/23/09 Angus Rodgers
1/23/09 Passerby
1/23/09 Dave Dodson
1/24/09 Albert
1/24/09 Angus Rodgers
1/24/09 Albert
1/26/09 Albert
1/26/09 Driveby
1/26/09 Albert
1/26/09 A N Niel
1/27/09 Albert