# What does -sin(arccos(2))+tan("arccsc"(4)) equal?

Jun 14, 2018

$- \sin \left(\arccos \left(2\right)\right) + \tan \left(\text{arccsc} \left(4\right)\right) = \frac{1}{15} \sqrt{15} - \sqrt{3} i$

#### Explanation:

We use the following identities to evaluate the above expression:

$\sin x = \sqrt{1 - {\cos}^{2} x}$

$\cot x = \sqrt{{\csc}^{2} x - 1}$

$\tan x = \frac{1}{\cot} x$

So

$- \sin \left(\arccos \left(2\right)\right) + \tan \left(\text{arccsc} \left(4\right)\right)$

$= - \sqrt{1 - {\cos}^{2} \left(\arccos 2\right)} + \frac{1}{\sqrt{{\csc}^{2} \left(\text{arccsc} 4\right) - 1}}$

$= - \sqrt{1 - 4} + \frac{1}{\sqrt{16 - 1}}$

$- \sqrt{- 3} + \frac{1}{\sqrt{15}}$

$= \frac{1}{15} \sqrt{15} - \sqrt{3} i$