Question

1+SecA-TanA/1+SecA+TanA=1-SinA/CosA (Hint sec^2A-Tan^2A=1)

Answer 1

L.H.S = 1 + secA – tanA / 1 + secA + tanA        
R.H.S =1 – sinA / cosA 
L.H.S=(sec2A – tan2A) + secA – tanA / 1 + secA + tanAL. H. S= (secA – tanA) (secA + tanA) + (secA – tanA) / 1 + secA + tanA

                                                                            [a2+b2=(a+b) (a-b)]
L.H.S = (secA – tanA) (1+secA + tanA) / 1 + secA + tanA
L.H.S = secA – tanA
                                                           ,[secA = 1 / cosA], [ tanA = sinA / cosA]
 putting the value of secA and tanA
= 1 / cosA - sinA / cosA
L.H. S= 1 – sinA / cosA
So here L.H. S is equal to R.H.S

Answer 2

I have solved this question. hope you will understand