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

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durbanville's profile pic

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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

 

 

Sources:
oldnick's profile pic

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`cos2A= 1/sqrt(1+tan^2 2A)`

`sin2A=(tan2A)/sqrt(1+tan^2 2A)`

subsituing in   `(cos2A)/(1+sin2A)` :

`(1/sqrt(1+tan^2 2A))/(1+(tan 2A)/(sqrt(1+tan^2 2A)))` `=1/(sqrt(1+tan^2 2A)+tan2A)`

multipling under and down ratio by `sqrt(1+tan^2 2A)-tan2A`

`1/(sqrt(1+tan^2 2A)+tan2A)=` `(sqrt(1+tan^2 2A)-tan2A)/(1+tan^2 2A - tan^2 2A)=` `=sqrt(1+tan^2 2A)-tan2A=`

`=sqrt(1+((2tanA)/(1-tan^2 A))^2)-(2tanA)/(1-tan^2 2A)=`

`=sqrt(1+(4tan^2 A)/(1-2tan^2 A +tan^4 A))-(2tan A)/(1-tan^2 A)=`

`=sqrt((1-2tan^2 A+tan^4 A+4tan^2 A)/((1-tan^2 A)^2))-(2tanA)/(1-tan^2 A)=`

`=sqrt((1+2tan^2 A+tan^4 A)/(1-tan^2 A)^2)-(2tanA)/(1-tan^2 A)=`

`=sqrt((1+tan^2A)^2/(1-tan^2 A)^2)-(2tanA)/(1-tan^2A)=`

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

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

`=(1-tanA)/(1+tanA)`

 

rockstar195699's profile pic

Posted on

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

    1.       =   (numerator)

     2.    (denominator)

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

1.        (numerator) and

2.       (denominator)

Now   1.  is divided by   2.

because the cos As will cancel out (divide )

Now multiply our new expression  by because it will allow us to simplify effectively to obtain the RHS and is the same as .

 

We know from double angle identities that :

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

Now rearrange the order of the square:

We know from identities that

and

Now put all the information of our denominator together:

Now combine with the numerator which we recall as:

therefore LHS=RHS

Thanks for the help I really appreciate it!!!!

I'm also from South Africa 

 

 

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