# How do you prove  tan^2x-1 = 1+tanx ?

Feb 10, 2016

The given equation is not true!

#### Explanation:

As a counter example, consider $x = 0$
$\textcolor{w h i t e}{\text{XXX}} \tan \left(x = 0\right) = 0$
$\textcolor{w h i t e}{\text{XXX}} {\tan}^{2} \left(x = 0\right) = 0$

${\tan}^{2} \left(x = 0\right) - 1 = - 1 \ne + 1 = 1 + \tan \left(x = 0\right)$

Feb 10, 2016

It's not an identity, so it can't be proven.

However, the equation can be solved and has the solutions $x = \arctan \left(2\right)$ or $x = \arctan \left(- 1\right)$.

#### Explanation:

You can't prove this because it isn't an identity.

If it was an identity, you would have:

${\tan}^{2} x - 1 = \left(\tan x + 1\right) \left(\tan x - 1\right) \stackrel{\text{? }}{=} 1 + \tan x$

which could only be true if $\tan x - 1 = 1$ was true for every $x$.

This is certainly not the case.

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However, even though you can't prove this as an identity (valid for all $x$), you can solve the equation and find some $x$ as solutions.

Let's do this.

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

$\iff {\tan}^{2} x - \tan x - 2 = 0$

Substitute $y = \tan x$:

${y}^{2} - y - 2 = 0$

$y = 2 \text{ or } y = - 1$.
$\tan x = 2 \text{ or } \tan x = - 1$
$x = \arctan \left(2\right) \text{ or } x = \arctan \left(- 1\right)$