(sin0 - cos0)^(2) - cot0tan0 = -2sin0cos0
*(all 0's = theta symbol)
Formula
$\displaystyle (a-b)^2 =a^2+b^2 -2ab$
Hence
$\displaystyle LHS=sin^2(\theta) +cos^2 (\theta) -2sin(\theta)cos(\theta) - cot(\theta)\frac{1}{cot(\theta)} $
$\displaystyle
=1-2sin(\theta)cos(\theta) - 1 ...............(\text{using } sin^2x+ cos^2x=1)
$
$\displaystyle
= RHS
$
1. You are supposed to know that
$\displaystyle \cot(\theta)=\dfrac1{\tan(\theta)}$
$\displaystyle (\sin(\theta))^2+(\cos(\theta))^2 = 1$
2. Expand the bracket:
$\displaystyle (\sin(\theta))^2-2\cos(\theta) \cdot \sin(\theta)+(\cos(\theta))^2 - \dfrac1{\tan(\theta)} \cdot \tan(\theta) = $
Using the properties mentioned above you'll get:
$\displaystyle 1 - 2\cos(\theta) \cdot \sin(\theta) -1 = -2\cos(\theta) \cdot \sin(\theta)$
EDIT: Too late ...