Question

Help....
I've done using many ways but I still fail to prove that:
\[\frac{tan(x)+sec(x)-1}{tan(x)-sec(x)+1}=tan(x)+sec(x)\] .

Please help me............

Answer 1

multiply divide LHS by (tan x + sec x)..
leave one in numerator..simply with denominator..
you should get your ans..

Answer 2

did you do what @shubhamsrg did?
Or you could do which is much longer:
I write both sides in terms of cosine and sine
Did you try that?

Answer 3

use fractions

tan = sin/cos sec = 1/cos 1 = cos/cos

so the numerator is

\[\frac{\sin(x)}{\cos(x)} + \frac{1}{\cos(x)} - \frac{\cos(x)}{\cos(x)}\]

this can be simplified to


\[\frac{\sin(x) - \cos(x) + 1}{\cos(x)}\]

same approach for the denominator

then when dividing by a fraction take the reciprocal and multiply

Answer 4

LHS multiply by\[\frac{tan(x)+sec(x)-1}{tan(x)-sec(x)+1}\times\frac{tan(x)+sec(x)}{tan(x)+sec(x)}\]
like this?

Answer 5

(tanx +secx -(sec^2x-tan^2x))/(tanx-secx+1) =(tanx+secx)(1-secx+tanx)/((tanx-secx+1) =(tanx+secx) ..

Answer 6

yes.. @matricked has done it perfectly right..

Answer 7

Oh I get it...
\[\frac{tan(x)+sec(x)-1}{tan(x)-sec(x)+1}\times \frac{tan(x)+sec(x)}{tan(x)+sec(x)}=\\
\frac{(tan(x)+sec(x)-1)(tan(x)+sec(x))}{tan^2(x)-sec^2(x)+tan(x)+sec(x)}=\\
\frac{(tan(x)+sec(x)-1)(tan(x)+sec(x)}{tan(x)+sec(x)-1}=tan(x)+sec(x)\]

Answer 8

bingo!!