# Prove The Equation

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• May 29th 2010, 05:35 AM
cyt91
Prove The Equation
Given A+B+C=90

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

Thanks!(Bow)
• May 29th 2010, 05:58 AM
Grandad
Hello cyt91
Quote:

Originally Posted by cyt91
Given A+B+C=90

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

Thanks!(Bow)

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

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

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

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

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

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

$\Rightarrow \cot A + \cot B + \cot C = \cot A \cot B \cot C$
Grandad
• May 29th 2010, 06:13 AM
Soroban
Hello, cyt91!

Another approach . . .

Quote:

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

Prove that: . $\cot\!A+\cot\!B+\cot\!C\:=\:\cot\!A\cot\!B\cot\!C$

We need this identity: . $\cot(\alpha + \beta) \:=\:\frac{\cot\alpha\cot\beta - 1}{\cot\alpha + \cot\beta}$

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

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

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

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

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

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

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

Nice explanation, Grandad!
.
• May 29th 2010, 06:57 PM
cyt91
Thanks a lot. You guys have been helpful. Keep up the good work!(Rofl)