Question

1+sec(A)-tan (A)/1+sec(A)+tan(A)=1-sin(A)/cos(A)

Answer 1

Sol:-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]

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

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....Proved

Thank You!

Answer 2

Answer:

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