# How do I prove ((1-tan^2 theta) / (1+tan^2 theta))=cos(2 theta)?

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Is proving ((1-tan^2 theta) / (1+tan^2 theta))=cos 2 theta the same as prove ((1-tan^2 theta) / (1+tan^2 theta))=cos (2 theta)?

I can prove the first, but don't know if the parentheses cos(2 theta) in the second is the same.

Is proving ((1-tan^2 theta) / (1+tan^2 theta))=cos 2 theta the same as prove ((1-tan^2 theta) / (1+tan^2 theta))=cos (2 theta)?

I can prove the first, but don't know if the parentheses cos(2 theta) in the second is the same.

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Here are the identities that I used:

Now here's the actual proof:

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As proved below

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We seek to prove that:

# (1-tan^2 theta) / (1+tan^2 theta) -= cos(2 theta) #

We will require the following trigonometric definitions/identities:

# tan phi = sin phi / cos phi #

# sin^2 phi + cos^phi -=1 #

# cos 2phi -= cos^2 phi - sin^2 phi #

Consider the LHS of the given expression:

# LHS = (1-tan^2 theta) / (1+tan^2 theta) #

# \ \ \ \ \ \ \ \ = (1-(sin^2 theta)/(cos^2 theta)) / (1+(sin^2 theta)/(cos^2 theta)) #

# \ \ \ \ \ \ \ \ = ((cos^2 theta -sin^2 theta)/(cos^2 theta)) / ( (cos^2 theta+sin^2 theta)/(cos^2 theta)) #

# \ \ \ \ \ \ \ \ = (cos^2 theta -sin^2 theta)/(cos^2 theta+sin^2 theta) #

# \ \ \ \ \ \ \ \ = (cos 2theta)/(1) #

# \ \ \ \ \ \ \ \ = cos 2theta #

# \ \ \ \ \ \ \ \ = RHS \ \ \ \ # QED

Describe your changes (optional) 200