How do you prove tan(x + (pi/2)) = -cotx?

Apr 16, 2015

We can not simply use the tangent of a sum formula, because $\tan \left(\frac{\pi}{2}\right)$ does not exist. See other answer for using the sum formula.

So use
$\tan \theta = \sin \frac{\theta}{\cos} \theta$ and the sum formulas for sine and cosine.

$\tan \left(x + \frac{\pi}{2}\right) = \frac{\sin \left(x + \frac{\pi}{2}\right)}{\cos \left(x + \frac{\pi}{2}\right)}$
$= \frac{\sin x \cos \left(\frac{\pi}{2}\right) + \cos x \sin \left(\frac{\pi}{2}\right)}{\cos x \cos \left(\frac{\pi}{2}\right) - \sin x \sin \left(\frac{\pi}{2}\right)} = \frac{0 + \cos x}{0 - \sin x} = - \cos \frac{x}{\sin} x = - \cot x$

Oct 25, 2017

If we really want to use the sum formula for tangent, then we can. See below.

Explanation:

We cannot simply apply the sum formula as tan(x+pi/2) = (tanx+tan(pi/2))/(1-tanxtan(pi/2) because $\tan \left(\frac{\pi}{2}\right)$ does not exist.

But we can rewrite $x + \frac{\pi}{2} = \left(x + \frac{\pi}{4}\right) + \frac{\pi}{4}$ and apply the sun formula twice.

For any $a$ in the domain of the tangent function, we have:

$\tan \left(a + \frac{\pi}{4}\right) = \frac{\tan a + \tan \left(\frac{\pi}{4}\right)}{1 - \tan a \tan \left(\frac{\pi}{4}\right)}$

$= \frac{\tan a + 1}{1 - \tan a}$

Therefore,
$\tan \left(x + \frac{\pi}{2}\right) = \tan \left(\left(x + \frac{\pi}{4}\right) + \frac{\pi}{4}\right)$

$= \frac{\tan \left(x + \frac{\pi}{4}\right) + 1}{1 - \tan \left(x + \frac{\pi}{4}\right)}$

$= \frac{\left[\frac{\tan x + 1}{1 - \tan x}\right] + 1}{1 - \left[\frac{\tan x + 1}{1 - \tan x}\right]}$

$= \frac{\left[\frac{\tan x + 1}{1 - \tan x}\right] + 1}{1 - \left[\frac{\tan x + 1}{1 - \tan x}\right]} \cdot \frac{1 - \tan x}{1 - \tan x}$

$= \frac{\tan x + 1 + 1 - \tan x}{1 - \tan x - \left(\tan x + 1\right)}$

$= \frac{2}{- 2 \tan x}$

$= - \cot x$