# How do you verify the identity (csc x - sin x)(sec x - cos x)(tan x + cot x) = 0?

Apr 23, 2015

I am not sure that is equal to zero....I got $1$!

Have a look.

Apr 23, 2015

WARNING: This may not be the simplest way!
$\left(\csc - \sin\right) \left(\sec - \cos\right) \left(\tan + \cot\right)$
Let $s$ represent $\sin \left(x\right)$
$c$ represent $\cos \left(x\right)$
and $t$ represent $\tan \left(x\right)$

$\left(\csc - \sin\right) \left(\sec - \cos\right) \left(\tan + \cot\right)$
becomes
$\left(\frac{1}{s} - s\right) \left(\frac{1}{c} - c\right) \left(t + \frac{1}{t}\right)$

$= \left(\frac{1}{s c} - t - \frac{1}{t} + s c\right) \cdot \left(t + \frac{1}{t}\right)$

((xx,|1/(sc),,-t,,-1/t,,+sc),( - , - , - , - , - , - , - , - - ),(1/t,|1/(sct),,-1,,-1/t^2,,+(sc)/t),(+t,|t/(sc),,-t^2,,-1,,sct) )

Replacing $t$ with $\frac{s}{c}$

we get
$\frac{1}{{s}^{2}} - 1 - {c}^{2} / {s}^{2} + {c}^{2} + \frac{1}{c} ^ 2 - {s}^{2} / {c}^{2} - 1 + {s}^{2}$

$= \frac{1 - {s}^{2}}{{c}^{2}} - 2 + \left({c}^{2} + {s}^{2}\right) + \frac{1 - {c}^{2}}{s} ^ 2$

$= 1$