Question

secA+tanA -1/tan A -secA +1 =cos/1 - sinA

Answer 1

multiply numerator and denominator by cos A

(1 + sin A  - cosA) / (sin A + cos A -1 )
multiply numerator and denominator by (sinA + cos A +1)

(1+sinA - cos A) (1+sinA+cosA)  / [ (sinA+ cosA) - 1 ] [ (sin A+cosA)+1]
[ (1+sin A)² - cos² A ] / [ (sinA+ cosA)² - 1² ]
[ 1 + sin²A + 2 sinA - cos²A ]  / [ sin²A+cos²A+2sinA cosA - 1 ]
= [ 2 sin² A + 2 sin A ]  /  2 sin A cos A
= 2 sin A [sin A + 1 ] / 2 sin A cos A
= (1 + sin A )/ cos A 
multiply by 1 - sine A in numerator , denomninator
= (1- sine ² A) / cos A  (1-sinA)
= cos²A / cosA (1-sinA)
cosA/1-sin A

Answer 2

Consider LHS
$\frac{secA+tanA -1}{tan A -secA +1}$

$\frac{(SecA+TanA) -(Sec^2A - Tan^2A)}{tan A -secA +1}$ [Replacing 1 with $Sec^2A - Tan^2A$]

$\frac{(SecA+TanA)(1-SecA + TanA)}{tan A -secA +1}$ [Cancelling (1-SecA + TanA) in numerator and denominator]

$SecA+TanA$   

$\frac{1}{CosA} + \frac{SinA}{CosA}$

$\frac{1+SinA}{CosA}$

$\frac{(1+SinA)(1-SinA)}{(CosA)(1-SinA)}$    [Multiplying numerator and denominator by (1-SinA)]

$\frac{(1-Sin^2A)}{(CosA)(1-SinA)}$ 

$\frac{(Cos^2A)}{(CosA)(1-SinA)}$

$\frac{CosA}{1-SinA}$