# How to prove the identity ? 1- (tan^2x/(1+tan^2x)) = cos^2x

Apr 3, 2018

$1 - \left({\tan}^{2} \frac{x}{1 + {\tan}^{2} x}\right) = {\cos}^{2} x$ color(white)(ddddwwwdd$\left[\text{as } \textcolor{red}{\tan x = \sin \frac{x}{\cos} x}\right]$

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

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

$1 - \left({\sin}^{2} \frac{x}{1}\right) = {\cos}^{2} x$ color(white)(ddddwwwdd$\left[\text{as } \textcolor{red}{{\cos}^{2} x + {\sin}^{2} x = 1}\right]$

$1 - {\sin}^{2} x = {\cos}^{2} x$

${\cos}^{2} x = {\cos}^{2} x$ color(white)(ddddwwwwwwwwdd$\left[\text{as } \textcolor{red}{{\cos}^{2} x = 1 - {\sin}^{2} x}\right]$

Hence Proved !

Apr 3, 2018
1. $1 + {\tan}^{2} x = {\sec}^{2} x$
2. $1 - \left({\tan}^{2} \frac{x}{\sec} ^ 2 x\right) = {\cos}^{2} x$
3. ${\tan}^{2} \frac{x}{\sec} ^ 2 x = {\sin}^{2} x$
4. $1 - {\sin}^{2} x = {\cos}^{2} x$

#### Explanation:

Line 1 is a common Pythagorean identity. Substituting gives line 2. Simplifying the complex fraction in 2 into sin and cos functions gives line 3. Line 4 is also a common Pythagorean identity.

Apr 3, 2018

Here is how I proved the identity:

#### Explanation:

$1 - \frac{{\tan}^{2} x}{1 + {\tan}^{2} x} = {\cos}^{2} x$

To solve this, we will use a bunch of trigonometric identities.

First, we know that ${\tan}^{2} x = {\sin}^{2} \frac{x}{\cos} ^ 2 x$ from quotient identities.
We also know that $1 + {\tan}^{2} x = {\sec}^{2} x$ from the pythagorean identities.

From these identities, we can rewrite the equation as:
$1 - \frac{{\sin}^{2} \frac{x}{\cos} ^ 2 x}{{\sec}^{2} x} = {\cos}^{2} x$

We also know that ${\sec}^{2} x = \frac{1}{\cos} ^ 2 x$ from reciprocal functions:
$1 - \frac{{\sin}^{2} \frac{x}{\cos} ^ 2 x}{\frac{1}{\cos} ^ 2 x} = {\cos}^{2} x$

Let's rewrite the division part with $\div$ instead of $/$ so it is easier to read:
$1 - \left({\sin}^{2} \frac{x}{\cos} ^ 2 x\right) \div \left(\frac{1}{\cos} ^ 2 x\right) = {\cos}^{2} x$

We know that dividing something is the same as multiplying it by the reciprocal, or $1$ over the number:
$1 - \left({\sin}^{2} \frac{x}{\cos} ^ 2 x\right) \cdot \left({\cos}^{2} \frac{x}{1}\right) = {\cos}^{2} x$

Since ${\cos}^{2} x$ is being both divided and multiplied, we can cross both of them out:
$1 - \left({\sin}^{2} \frac{x}{\cancel{{\cos}^{2} x}}\right) \cdot \left(\frac{\cancel{{\cos}^{2} x}}{1}\right) = {\cos}^{2} x$

And now the cleaned up version looks like this:
$1 - {\sin}^{2} x = {\cos}^{2} x$

Finally, we know that $1 - {\sin}^{2} x = {\cos}^{2} x$ from the pythagorean identities:
${\cos}^{2} x = {\cos}^{2} x$

Hope this helps!