# How do you verify  tan (x+pi/2)= -cot x?

Feb 16, 2016

By applying the basic trigonometric relations. see below

#### Explanation:

Key-relation 1. $\tan x = \sin \frac{x}{\cos} x$

Key-relation 2. $\cot x = \frac{1}{\tan} x = \cos \frac{x}{\sin} x$

Key-relation 3. $\cos \left(a + b\right) = \cos a \cdot \cos b - \sin a \cdot \sin b$

Key-relation 4. $\sin \left(a + b\right) = \cos a \cdot \sin b + \sin a \cdot \cos b$

Import results

Important result 1. $\cos \left(\frac{\pi}{2}\right) = 0$
Important result 1. $\sin \left(\frac{\pi}{2}\right) = 1$

Gathering

By using all the knowledge left so far and some mathematical tricks, we have:

$\sin \left(x + \left(\frac{\pi}{2}\right)\right) = \cos x \cdot \sin \left(\frac{\pi}{2}\right) + \sin x \cdot \cos \left(\frac{\pi}{2}\right)$
$\sin \left(x + \left(\frac{\pi}{2}\right)\right) = \cos x$

Further:

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

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

Finally:

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

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

End of the proof!