# Thread: [SOLVED] Prove that this is true:

1. ## [SOLVED] Prove that this is true:

(sin0 - cos0)^(2) - cot0tan0 = -2sin0cos0

*(all 0's = theta symbol)

2. Originally Posted by moonman
(sin0 - cos0)^(2) - cot0tan0 = -2sin0cos0

*(all 0's = theta symbol)
Formula
$\displaystyle (a-b)^2 =a^2+b^2 -2ab$

Hence
$\displaystyle LHS=sin^2(\theta) +cos^2 (\theta) -2sin(\theta)cos(\theta) - cot(\theta)\frac{1}{cot(\theta)}$

$\displaystyle =1-2sin(\theta)cos(\theta) - 1 ...............(\text{using } sin^2x+ cos^2x=1)$

$\displaystyle = RHS$

3. Originally Posted by moonman
(sin0 - cos0)^(2) - cot0tan0 = -2sin0cos0

*(all 0's = theta symbol)
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 ...