# Tangents

• Jan 28th 2010, 01:38 AM
Punch
Tangents
If $\displaystyle A+B+C=90^\circ$, show that $\displaystyle tanAtanB+tanBtanC+tanCtanA=1$
• Jan 28th 2010, 04:47 AM
Hello Punch
Quote:

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

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

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

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

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

I think you can finish up now.

• Jan 28th 2010, 11:58 PM
Punch
Thanks but I don't understand how to relate them..
• Jan 29th 2010, 12:36 AM
Hello Punch

Here's the last bit, then:

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

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

• Jan 30th 2010, 06:12 AM
Punch
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 AM
Hello Punch
Quote:

Originally Posted by Punch
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?

I'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:
$\displaystyle A+B+C=90^o$
$\displaystyle \Rightarrow A+B = 90^o-C$

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

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

$\displaystyle \Rightarrow \tan A\tan B + \tan B\tan C +\tan C \tan A = 1$
So we have now done everything we were asked to do. Are there any steps in the working that you don't understand?

• Jan 30th 2010, 07:18 PM
Punch
I don't understand $\displaystyle \Rightarrow \tan A\tan B + \tan B\tan C +\tan C \tan A = 1$
• Jan 30th 2010, 07:24 PM
Prove It
Expand the brackets and then add $\displaystyle \tan{A}\tan{B}$ to both sides...
• Feb 1st 2010, 05:03 AM
Punch
To solve this question, I have to relate it to the tangent additional formulae.
• Feb 1st 2010, 06:03 AM
Hello Punch
Quote:

Originally Posted by Punch
To solve this question, I have to relate it to the tangent additional formulae.

Yes, and you'll see that I have used the tangent addition formula where $\displaystyle \tan(A+B)$ is written on the following line as $\displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}$.

• Feb 3rd 2010, 04:05 AM
Punch
Quote:

Hello PunchYes, and you'll see that I have used the tangent addition formula where $\displaystyle \tan(A+B)$ is written on the following line as $\displaystyle \frac{\tan A + \tan B}{1-\tan A \tan B}$.

I am sorry to be posting on an "old" thread. However, I was wondering how I could know when to relate such questions to formulas. (Wondering)(Wondering)(Wondering)(Wondering)

• Feb 3rd 2010, 04:24 AM
Hello Punch
Quote:

Originally Posted by Punch
I am sorry to be posting on an "old" thread. However, I was wondering how I could know when to relate such questions to formulas. (Wondering)(Wondering)(Wondering)(Wondering)