Question

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

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 + tanA

As we know that [sec2A – tan2A = 1]

So here L. H. S= (secA – tanA) (secA + tanA) + (secA – tanA) / 1 + secA + tanA

We know about this formula [a2+b2=(a+b) (a-b)]

L.H.S = (secA – tanA) (1+secA + tanA) / 1 + secA + tanA

L.H.S = secA – tanA

We know about the formula of secA and tanA,[secA = 1 / cosA], [ tanA = sinA / cosA]

 putting the value of secA and tanA

so = 1 / cosA - sinA / cosA

L.H. S= 1 – sinA / cosA

So here L.H. S is equal to R.H.S

so I hope it will help u..

mark brainest if it helped