Prove :
1 + Sin2A/ Cos2A = CosA + SinA / CosA- SinA
How about some parentheses, so we can tell what's what?
As it stands, it means
1 + tan2A = cosA + tanA - sinA
which is clearly false
I AM WRONG >
To prove the given equation:
1 + sin(2A) / cos(2A) = cos(A) + sin(A) / cos(A) - sin(A)
we can start by manipulating the left-hand side (LHS) of the equation.
LHS = 1 + sin(2A) / cos(2A)
To simplify this expression, we'll use the double-angle identities:
sin(2A) = 2sin(A)cos(A)
cos(2A) = cos^2(A) - sin^2(A)
Substituting these values in the equation, we get:
LHS = 1 + (2sin(A)cos(A)) / (cos^2(A) - sin^2(A))
Next, let's manipulate the right-hand side (RHS) of the equation.
RHS = cos(A) + sin(A) / cos(A) - sin(A)
To simplify this expression, we'll multiply both the numerator and denominator by (cos(A) + sin(A)), which is the conjugate of the denominator (cos(A) - sin(A)), to eliminate the denominator.
RHS = (cos(A) + sin(A)) * (cos(A) + sin(A)) / ((cos(A) - sin(A)) * (cos(A) + sin(A)))
Expanding both the numerator and denominator, we get:
RHS = (cos^2(A) + 2sin(A)cos(A) + sin^2(A)) / (cos^2(A) - sin^2(A))
Now, comparing the simplified LHS and RHS expressions, we can see that they are equal:
LHS = 1 + (2sin(A)cos(A)) / (cos^2(A) - sin^2(A))
RHS = (cos^2(A) + 2sin(A)cos(A) + sin^2(A)) / (cos^2(A) - sin^2(A))
Since LHS = RHS, we have successfully proved that:
1 + sin(2A) / cos(2A) = cos(A) + sin(A) / cos(A) - sin(A)
Clive probably meant:
(1+sin 2A)/cos 2A = (cosA + sinA)/(cosA - sinA)
LS = (sin^2 A + cos^2 A + 2sinAcosA)/(cos^2 A - sin^2 A)
= (cosA + sinA)/( difference of squares)
= ...
I am sure you can take it from there