# Prove the double angle identity: `(1-tanA)/(1+tan A) = (cos 2A)/(1+sin 2A)` : 1-tan A       cos2A ---------- = ---------- 1+tan A     1+sin2A

`(1-tan A)/(1+tanA)= (cos 2A)/(1+sin 2A)`

To prove that the LHS (left hand side) = RHS (right hand side) we must manipulate the expressions. Changing the left hand side

1.     `1-tan A`  = `1-sinA/cosA`  (numerator)

2.    `1+tan A = 1+sinA/cosA` (denominator)

Now use the LCD for 1.  and 2.  ...

`(1-tan A)/(1+tanA)= (cos 2A)/(1+sin 2A)`

To prove that the LHS (left hand side) = RHS (right hand side) we must manipulate the expressions. Changing the left hand side

1.     `1-tan A`  = `1-sinA/cosA`  (numerator)

2.    `1+tan A = 1+sinA/cosA` (denominator)

Now use the LCD for 1.  and 2.   which is `cosA`

1.   `therefore 1-sinA/cosA = (cosA - sinA)/cos A`     (numerator) and

2. `therefore 1+sinA/cosA = (cosA + sinA)/cos A`      (denominator)

Now   1.  is divided by   2.

`therefore ((cosA-sinA)/cosA) divide ((cosA+sinA)/cosA)`

`therefore =(cosA-sinA)/(cosA+sinA)` because the cos As will cancel out (divide )

Now multiply our new expression  by `(cosA +sin A)/(cosA+sinA)` because it will allow us to simplify effectively to obtain the RHS and is the same as `1/1` .

`therefore = (cosA-sinA)/(cosA+sinA) times (cosA+sinA)/(cosA+sinA)`

`therefore = (cos^2A - sin^2A)/(cosA + sinA)^2`

We know from double angle identities that :`cos^2A-sin^2=cos2A`

We know from the perfect square (our denominator) that:

`(cosA+sinA)^2 = cos^2A +2sinAcosA + sin^2A`

Now rearrange the order of the square:

`cos^2A +sin^2A + 2sinAcosA`

We know from identities that `cos^2A +sin^2A = 1`

`` and `2sinAcosA = sin2A`

Now put all the information of our denominator together:

`therefore (cosA+sinA)^2 = 1+sin2A`

Now combine with the numerator which we recall as:

`sin^2A-cos^2a = cos 2A`

`therefore = (cos2A)/ (1+sin2A)`

therefore LHS=RHS

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