How do you verify the identity $(tanx - secx + 1)/(tanx + secx - 1) = cosx/(1 + sinx)$ ?

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Answer 1 · 2018-04-30T13:22:18

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

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

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

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

$=1/(secx+tanx)$

$=1/(1/cosx+sinx/cosx)$

$=1/((1+sinx)/cosx)$

$= cosx/(1 + sinx)=RHS$

Alternative method

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

$=(cos^2x(tanx - secx + 1))/(cos^2x(tanx + secx - 1))$

$=(cosx(sinx/cosx*cosx - 1/cos*cosx + cosx))/(cos^2x*sinx/cosx + cos^2x*1/cosx - cos^2x)$

$=(cosx(sinx- 1 + cosx))/(cosx*sinx + cosx - (1-sin^2x))$

$=(cosx(sinx- 1 + cosx))/(cosx(sinx + 1) - (1-sinx)(1+sinx))$

$=(cosx(sinx- 1 + cosx))/((1+sinx)(cosx - 1+sinx))$

$=cosx/(1+sinx) =RHS$

How do you verify the identity (tanx - secx + 1)/(tanx + secx - 1) = cosx/(1 + sinx)(tanx - secx + 1)/(tanx + secx - 1) = cosx/(1 + sinx)?