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

Jun 9, 2015

Verify tan (x + pi/2) = - cot x

#### Explanation:

On the trig unit circle, the value AT = tan x rotates counterclockwise an arc of pi/2, and becomes BU = - cot x

Jun 24, 2016

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

For the numerator, use $\sin \left(A + B\right) = \sin \left(A\right) \cos \left(B\right) + \cos \left(A\right) \sin \left(B\right)$:

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

$= \sin \left(x\right) \cdot 0 + \cos \left(x\right) \cdot 1$

$= \cos \left(x\right)$

In the denominator, use $\cos \left(A + B\right) = \cos \left(A\right) \cos \left(B\right) - \sin \left(A\right) \sin \left(B\right)$:

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

$= \cos \left(x\right) \cdot 0 - \sin \left(x\right) \cdot 1$

$= - \sin \left(x\right)$

Thus, we see that

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

$\square$