# How do you prove the identity tan^4x+tan^2x+1=(1-cos^2x)(sin^2x)/(cos^4x)?

Sep 22, 2015

You don't, because it's false.

#### Explanation:

${\tan}^{4} \left(x\right) + {\tan}^{2} \left(x\right) + 1 = \left(1 - {\cos}^{2} \left(x\right)\right) {\sin}^{2} \frac{x}{\cos} ^ 4 \left(x\right)$

We know the Pythagorean identity that ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

So we know that ${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$, rewriting that

${\tan}^{4} \left(x\right) + {\tan}^{2} \left(x\right) + 1 = {\sin}^{4} \frac{x}{\cos} ^ 4 \left(x\right)$

Since $\sin \frac{x}{\cos} \left(x\right) = \tan \left(x\right)$ we can rewrite the right side as ${\tan}^{4} \left(x\right)$

${\tan}^{4} \left(x\right) + {\tan}^{2} \left(x\right) + 1 = {\tan}^{4} \left(x\right)$

Cancelling the ${\tan}^{4} \left(x\right)$ from both sides,
${\tan}^{2} \left(x\right) + 1 = 0$

Or

${\tan}^{2} \left(x\right) = - 1$

Which doesn't have any solutions, therefore the equation is false for all values of x.