[SOLVED] Prove that this is true:

**moonman** · February 19th 2009, 11:11 PM

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

*(all 0's = theta symbol)

**ADARSH** · February 19th 2009, 11:24 PM

Originally Posted by moonman

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

*(all 0's = theta symbol)

Formula
$(a-b)^2 =a^2+b^2 -2ab$

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

$<br /> =1-2sin(\theta)cos(\theta) - 1 ...............(\text{using } sin^2x+ cos^2x=1)<br />$

$<br /> = RHS <br />$

**earboth** · February 19th 2009, 11:26 PM

Originally Posted by moonman

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

*(all 0's = theta symbol)

1. You are supposed to know that

$\cot(\theta)=\dfrac1{\tan(\theta)}$

$(\sin(\theta))^2+(\cos(\theta))^2 = 1$

2. Expand the bracket:

$(\sin(\theta))^2-2\cos(\theta) \cdot \sin(\theta)+(\cos(\theta))^2 - \dfrac1{\tan(\theta)} \cdot \tan(\theta) =$

Using the properties mentioned above you'll get:

$1 - 2\cos(\theta) \cdot \sin(\theta) -1 = -2\cos(\theta) \cdot \sin(\theta)$

EDIT: Too late ...

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