`(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. ...
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`(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