Question

(secA+tanA-1)/(tanA-secA+1)=(cosA)/(1-sinA)

Answer 1

(secA+tanA-(sec^2 A - tan ^2A)) )/tanA-secA+1) as sec^2 A= 1 + tan ^2 A so 1= sec^2 A - tan ^2 A 
= (sec A + tan A - (sec A + tan A) ( sec A - tan A)) / ( tanA-secA+1) 
= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA-secA+1) 
= (sec A + tan A)/(+ tan A - sec A)/(tan A- sec A+ 1) 
= (sec A + tan A) 
= 1/cos A + sin A/ cos A 
= (1+ sin A)/ cos A 
= (1 + sin A )(1- sin A)/(cos A (1- sin A)) 
= (1- sin ^2 A/(cos A (1- sin A)) 
= cos ^2 A / (cos A (1- sin A)) 
= cos A /(1- sin A) 

proved

Answer 2

secA+tanA-1)/tanA-secA+1)

We know that sec^2-tan^2=1

=(secA+tanA-(sec^2 A - tan ^2A)) /tanA-secA+1)

We also know that a^2-b^2=(a+b)(a-b)

= (sec A + tan A - (sec A + tan A) ( sec A - tan A)) / ( tanA-secA+1)

= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA-secA+1)
= (sec A + tan A)/(+ tan A - sec A)/(tan A- sec A+ 1)
= (sec A + tan A)

= 1/cos A + sin A/ cos A
= (1+ sin A)/ cos A
= (1 + sin A )(1- sin A)/(cos A (1- sin A))
= (1- sin ^2 A/(cos A (1- sin A))
= cos ^2 A / (cos A (1- sin A))
= cos A /(1- sin A)

Hence proved