# What does cos(arctan(pi/2))-sin(arc cot(pi/4))  equal?

Jun 8, 2016

$\cos \left(a r c \tan \left(\frac{\pi}{2}\right)\right) - \sin \left(a r c \cot \left(\frac{\pi}{4}\right)\right) = \frac{2}{\sqrt{{\pi}^{2} + 4}} - \frac{4}{\sqrt{{\pi}^{2} + 16}}$

#### Explanation:

$\cos \left(\arctan \left(\frac{\pi}{2}\right)\right) - \sin \left(a r c \cot \left(\frac{\pi}{4}\right)\right)$
We know,
$\cos \left(\arctan \left(\frac{\pi}{2}\right)\right)$
using the following identity,
$\cos \left(\arctan \left(x\right)\right) = \frac{1}{\sqrt{1 + {x}^{2}}}$
$= \frac{1}{\sqrt{1 + {\left(\frac{\pi}{2}\right)}^{2}}}$
Refining it,
$= \setminus \frac{2}{\setminus \sqrt{4 + \setminus {\pi}^{2}}}$

Also,
$\sin \left(a r c \cot \left(\frac{\pi}{4}\right)\right)$
using the following identity,
$\sin \left(a r c \cot \left(x\right)\right) = \frac{1}{\sqrt{1 + {x}^{2}}}$
$\sin \left(a r c \cot \left(x\right)\right) = \frac{1}{\sqrt{1 + {x}^{2}}}$
Refining it,
$= \frac{4}{\sqrt{16 + {\pi}^{2}}}$

Finally,
$= \frac{2}{\sqrt{{\pi}^{2} + 4}} - \frac{4}{\sqrt{{\pi}^{2} + 16}}$