$\displaystyle \frac{(tan \theta + sec \theta -1 )}{(tan \theta - sec \theta +1 )}= tan \theta + sec \theta$

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- Jun 23rd 2009, 06:17 AMz1llchidentity
$\displaystyle \frac{(tan \theta + sec \theta -1 )}{(tan \theta - sec \theta +1 )}= tan \theta + sec \theta$

- Jun 23rd 2009, 06:28 AMalexmahone
The identity can be rewritten as:

$\displaystyle tan \theta+sec \theta - 1=(tan \theta-sec \theta+1)(tan\theta+sec\theta)$

$\displaystyle tan \theta+sec \theta - 1=tan ^2\theta -sec \theta tan \theta + tan \theta + sec \theta tan\theta - sec^2\theta+sec \theta$

$\displaystyle tan \theta+sec \theta - 1=tan^2\theta +tan\theta - sec^2\theta+sec \theta$

$\displaystyle tan \theta+sec \theta -1=-1+tan\theta+sec\theta$

which is always true. - Jun 23rd 2009, 06:28 AMz1llch
arent i suppose to use the left hand side only, and prove it is equal to the right hand side?

- Jun 23rd 2009, 07:07 AMalexmahone
- Jun 23rd 2009, 07:13 AMz1llch
hahaha. i'm stuck with the number 1. guessed i'm supposed to use the cos2+sin2=1 formula to get rid of it, but do not know how. or maybe i;m wrong, there other way. (Nerd)

- Jun 23rd 2009, 08:13 AMgreat_math
We can write it as,

$\displaystyle \frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}$

$\displaystyle = \frac{\tan\theta+\sec\theta+\tan^2\theta-\sec^2\theta}{\tan\theta-\sec\theta+1}$

$\displaystyle = \frac{(\tan\theta+\sec\theta)(\tan\theta-\sec\theta+1)}{\tan\theta-\sec\theta+1}$

$\displaystyle = \tan\theta+\sec\theta$ - Jun 23rd 2009, 09:40 AMJhevon
- Jun 23rd 2009, 11:54 PMz1llch
thank you great_math.

took me an hour to come up with those lines this morning. but mine is way longer.. (Whew)