1. ## Prove The Equation

Given A+B+C=90

Prove that cotA+cotB+cotC=(cotA)(cotB)(cotC)

Thanks!

2. Hello cyt91
Originally Posted by cyt91
Given A+B+C=90

Prove that cotA+cotB+cotC=(cotA)(cotB)(cotC)

Thanks!
Note first that $\displaystyle \cot\theta = \tan(90^o-\theta)$

So:
$\displaystyle A+B+C = 90^o$

$\displaystyle \Rightarrow A = 90^o-(B+C)$

$\displaystyle \Rightarrow \cot A=\tan(B+C)$, using the above result
$\displaystyle =\frac{\tan B + \tan C}{1-\tan B \tan C}$

$\displaystyle =\frac{\cot C+\cot B}{\cot B \cot C - 1}$, multiplying top-and-bottom by $\displaystyle \cot B \cot C$

$\displaystyle \Rightarrow \cot A \cot B \cot C - \cot A = \cot B + \cot C$

$\displaystyle \Rightarrow \cot A + \cot B + \cot C = \cot A \cot B \cot C$

3. Hello, cyt91!

Another approach . . .

Given: .$\displaystyle A+B+C\:=\:90^o$

Prove that: .$\displaystyle \cot\!A+\cot\!B+\cot\!C\:=\:\cot\!A\cot\!B\cot\!C$
We need this identity: .$\displaystyle \cot(\alpha + \beta) \:=\:\frac{\cot\alpha\cot\beta - 1}{\cot\alpha + \cot\beta}$

Also note that: .$\displaystyle \cot(90^o - \alpha) \;=\;\tan\alpha \;=\;\frac{1}{\cot\alpha}$

We have: - - . $\displaystyle A + B \;=\;90^o - C$

Then: . . . $\displaystyle \cot(A + B) \;=\;\cot(90^o - C)$

. . . . . .$\displaystyle \frac{\cot\!A\cot\!B - 1}{\cot\!A + \cot\!B} \;=\;\frac{1}{\cot\!C}$

. .$\displaystyle \cot\!C(\cot\!A\cot\!B - 1) \;=\;\cot\!A + \cot\!B$

$\displaystyle \cot\!A\cot\!B\cot\!C - \cot\!C \;=\;\cot\!A+ \cot\!B$

. .$\displaystyle \cot\!A + \cot\!B + \cot\!C \;=\;\cot\!A\cot\!B\cot\!C$