Question

sinA-CosA+1/sinA+CosA-1. = cosA/1-SinAa​

Answer 1

Answer:

LHS = sinA-cosA+1/sinA+cosA-1

divide both numerator and denominator by cosA

LHS = (tanA - 1 + secA) / (tanA + 1 - secA)

Now

sec^2 A = 1 + tan^2 A

sec^2 A - tan^2 A = 1

Using above relation at denominator of LHS

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

LHS = (tanA - 1 + secA) / ((secA - tanA) (-1 + secA + tanA)

LHS = 1 / (secA-tanA)

LHS = RHS

Hence Proved.

I think above proof will clear your doubt,

Answer 2

Answer:

thanks for the free point