Show that ((1+tanx-secx)/(secx+tanx-1))= (1+secx-tanx)/(secx+tanx+1)?

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Answer 1 · 2018-06-27T08:54:11

Yes

Let

$u = 1 + sec x + tan x$ $u=1+secx+tanx$ , $w = tan x - sec x$ $w=tanx-secx$

$\frac{1 + w}{u - 2} = \frac{1 - w}{u}$ $(1+w)/(u-2)=(1-w)/u$

$cancelu+uw=cancelu-uw+2w-2$

$uw=w-1$

Substitute

$(1+secx+tanx)(tanx-secx)=tanx-secx-1$

$canceltanxcancel(-secx)cancel(+secxtanx)-sec^2x+tan^2xcancel(-secxtanx)=canceltanxcancel(-secx)-1$

$sec^2x-tan^2x=1$

$:.$

Answer 2 · 2018-06-27T14:11:49

We know

$sec^2x-tan^2x=1$

$=>(secx-tanx)(secx+tanx)=1$

$=>(secx+tanx)/1=1/(secx-tanx)$

By componendo and dividendo

$=>(secx+tanx+1)/(secx+tanx-1)=(1+secx-tanx)/(1-secx+tanx)$

$=>(1+tanx-secx)/(secx+tanx-1)=(1+secx-tanx)/(secx+tanx+1)$