If $\displaystyle A+B+C=90^\circ$, show that $\displaystyle tanAtanB+tanBtanC+tanCtanA=1$

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- Jan 28th 2010, 01:38 AMPunchTangents
If $\displaystyle A+B+C=90^\circ$, show that $\displaystyle tanAtanB+tanBtanC+tanCtanA=1$

- Jan 28th 2010, 04:47 AMGrandad
- Jan 28th 2010, 11:58 PMPunch
Thanks but I don't understand how to relate them..

- Jan 29th 2010, 12:36 AMGrandad
Hello Punch

Here's the last bit, then:

$\displaystyle \frac{\tan A + \tan B}{1-\tan A\tan B}= \cot C$$\displaystyle \Rightarrow \tan C(\tan A + \tan B) = 1 - \tan A \tan B$$\displaystyle =\frac{1}{\tan C}$

$\displaystyle \Rightarrow \tan A\tan B + \tan B\tan C +\tan C \tan A = 1$

Grandad - Jan 30th 2010, 06:12 AMPunch
It is all so confusing... I have tried to combine the 2 posts but it makes no sense to me.

If $\displaystyle A+B+C=90^\circ$, show that $\displaystyle tanAtanB+tanBtanC+tanCtanA=1$

How about telling me the idea of solving this question? Is it about changing everything in terms of A? - Jan 30th 2010, 10:22 AMGrandad
Hello PunchI'm not sure what you mean by this. You were asked to show that if a certain statement was true (namely, $\displaystyle A+B+C=90^o$) then another statement follows (namely, $\displaystyle \tan A\tan B+\tan B\tan C+\tan C\tan A=1$).

I have done this by starting with the equation $\displaystyle A+B+C=90^o$ and ending with the equation $\displaystyle \tan A\tan B+\tan B\tan C+\tan C\tan A=1$.

Putting together my two posts, we get:$\displaystyle A+B+C=90^o$$\displaystyle \Rightarrow A+B = 90^o-C$So we have now done everything we were asked to do. Are there any steps in the working that you don't understand?

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

$\displaystyle \Rightarrow \frac{\tan A + \tan B}{1-\tan A\tan B}= \cot C$$\displaystyle \Rightarrow \tan C(\tan A + \tan B) = 1 - \tan A \tan B$$\displaystyle =\frac{1}{\tan C}$

$\displaystyle \Rightarrow \tan A\tan B + \tan B\tan C +\tan C \tan A = 1$

Grandad - Jan 30th 2010, 07:18 PMPunch
I don't understand $\displaystyle

\Rightarrow \tan A\tan B + \tan B\tan C +\tan C \tan A = 1

$ - Jan 30th 2010, 07:24 PMProve It
Expand the brackets and then add $\displaystyle \tan{A}\tan{B}$ to both sides...

- Feb 1st 2010, 05:03 AMPunch
To solve this question, I have to relate it to the tangent additional formulae.

- Feb 1st 2010, 06:03 AMGrandad
- Feb 3rd 2010, 04:05 AMPunch
- Feb 3rd 2010, 04:24 AMGrandad
Hello PunchThanks for the concession to my eyesight!

As far as your question is concerned, I'm afraid that experience - practice, practice, practice - is the only answer: just like learning to play a musical instrument.

There are just so many trig formulae, it's impossible to give any simple answer as to what to use and when. Since there are so many formulae, I always found it helpful myself - and so I taught my students - to recite the formulae aloud that I was trying to learn. Write them on Post-It stickers, and stick them up in places where you'll see them while you're doing something else - something that doesn't require much brain power (peeling potatoes, painting a fence, listening to your girlfriend, etc, etc) and recite them over and over. (Perhaps I wasn't really serious about the girlfriend bit...)

Grandad