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

## 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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Aviv S. Share
Feb 8, 2018

Here are the identities that I used:

${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

${\tan}^{2} \theta = {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta$

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

$\cos \left(2 \theta\right) = {\cos}^{2} \theta - {\sin}^{2} \theta$

Now here's the actual proof:

$\frac{1 - {\tan}^{2} \theta}{1 + {\tan}^{2} \theta} = \cos \left(2 \theta\right)$

$\frac{- {\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta + 1}{{\sin}^{2} \frac{\theta}{\cos} ^ 2 \theta + 1} = \cos \left(2 \theta\right)$

$\frac{\frac{- {\sin}^{2} \theta + {\cos}^{2} \theta}{\cos} ^ 2 \theta}{\frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\cos} ^ 2 \theta} = \cos \left(2 \theta\right)$

$\frac{- {\sin}^{2} \theta + {\cos}^{2} \theta}{\cancel{{\sin}^{2} \theta + {\cos}^{2} \theta}} = \cos \left(2 \theta\right)$

$- {\sin}^{2} \theta + {\cos}^{2} \theta = \cos \left(2 \theta\right)$

${\cos}^{2} \theta - {\sin}^{2} \theta = \cos \left(2 \theta\right)$

$\cos \left(2 \theta\right) = \cos \left(2 \theta\right)$

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Feb 10, 2018

As proved below

#### Explanation:

$1 + {\tan}^{2} \theta = {\sec}^{2} \theta = \frac{1}{\cos} ^ 2 \theta$

$1 - 2 {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1 = {\cos}^{2} \theta - {\sin}^{2} \theta = \cos \left(2 \theta\right)$

(1-tan^2theta) / (1+tan^2theta) = (1-(sin^2theta/cos^2theta))/(1/cos^2(theta) as $1 + {\tan}^{2} \theta = {\sec}^{2} \theta = \frac{1}{\cos} ^ 2 \theta$

$\left(\frac{{\cos}^{2} \theta - {\sin}^{2} \theta}{\cancel{{\cos}^{2} \left(\theta\right)}} / \frac{1}{\cancel{{\cos}^{2} \left(\theta\right)}}\right) = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \theta = \cos \left(2 \theta\right)$

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Steve M Share
Feb 8, 2018

We seek to prove that:

$\frac{1 - {\tan}^{2} \theta}{1 + {\tan}^{2} \theta} \equiv \cos \left(2 \theta\right)$

We will require the following trigonometric definitions/identities:

$\tan \phi = \sin \frac{\phi}{\cos} \phi$
${\sin}^{2} \phi + {\cos}^{\phi} \equiv 1$
$\cos 2 \phi \equiv {\cos}^{2} \phi - {\sin}^{2} \phi$

Consider the LHS of the given expression:

$L H S = \frac{1 - {\tan}^{2} \theta}{1 + {\tan}^{2} \theta}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1 - \frac{{\sin}^{2} \theta}{{\cos}^{2} \theta}}{1 + \frac{{\sin}^{2} \theta}{{\cos}^{2} \theta}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\frac{{\cos}^{2} \theta - {\sin}^{2} \theta}{{\cos}^{2} \theta}}{\frac{{\cos}^{2} \theta + {\sin}^{2} \theta}{{\cos}^{2} \theta}}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{{\cos}^{2} \theta - {\sin}^{2} \theta}{{\cos}^{2} \theta + {\sin}^{2} \theta}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\cos 2 \theta}{1}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \cos 2 \theta$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = R H S \setminus \setminus \setminus \setminus$ QED

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