proving trigo identities

**colloquial** · June 28th 2007, 07:44 AM

1. Given that A=B+C, prove that tanA-tanB-tanC=tanAtanBtanC

I can do it but if it's tan A + tan B...not minus.

and I have a real problem with proving. problem is, I don't know where to start from if I'm trying to solve a 'prove' question. any advice?

thank you so much!

**CaptainBlack** · June 28th 2007, 07:58 AM

Originally Posted by colloquial

1. Given that A=B+C, prove that tanA-tanB-tanC=tanAtanBtanC

I can do it but if it's tan A + tan B...not minus.

and I have a real problem with proving. problem is, I don't know where to start from if I'm trying to solve a 'prove' question. any advice?

thank you so much!

It's just a rewrite of the tan of a sum identity:

$ \tan(B+C) = \frac{\tan(B)+\tan(C)}{1-\tan(B)\tan(C)} $

so:

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

rearranging:

$ \tan(A) - \tan(A) \tan(B) \tan(C)= \tan(B)+\tan(C) $

then:

$ \tan(A) - \tan(B) -\tan(C)=\tan(A)\tan(B)\tan(C) $

Here we start from the known, and work towards what we have to prove.

RonL

**Soroban** · June 28th 2007, 08:03 AM

Hello, colloquial!

This is easier than you think . . .

1. Given that $A\:=\:B+C$ , prove that: . $\tan A -\tan B - \tan C \:=\:\tan A\tan B\tan C$

We have been given: . $A\;=\;B+C$

Take the tangent of both sides: . $\tan A \;=\;\tan(B + C)$

. . and we have: . $\tan A \;=\;\frac{\tan B + \tan C}{1 - \tan B\tan C}\quad\Rightarrow\quad \tan A \,-\, \tan A\tan B\tan C \;=\;\tan B \,+\,\tan C$

Therefore: . $\tan A - \tan B - \tan C \;=\;\tan A\tan B\tan C$

Edit: Too fast for me, Cap'n!

**CaptainBlack** · June 28th 2007, 08:13 AM

Originally Posted by Soroban

Edit: Too fast for me, Cap'n!

Hope this helps:

"I woke up this mornin'
with those beaten to MathHelpForum reply blues <der dar dur dum>..."

CFB

**Jhevon** · June 28th 2007, 08:14 AM

Originally Posted by CaptainBlack

CFB

is that an abbreviation of your blues name? what's it stand for?

**colloquial** · June 28th 2007, 08:16 AM

thank you!
oh dear lord I never expected it to be this easy, I went all the way:
expanding tanAtanBtanC to thinking that A+B+C=180 (although it wasn't stated).

proving: not my favourite.

**Soroban** · June 28th 2007, 08:47 AM

Hello, colloquial!

. . . thinking that $A+B+C\,=\,180$ (although it wasn't stated).

The other identity you referred to is one of my favorites.
. . I also started with angles $A,\,B,\,C$ in a triangle . . . *blush*

**CaptainBlack** · June 28th 2007, 12:38 PM

Originally Posted by Jhevon

is that an abbreviation of your blues name? what's it stand for?

See the Blues thread.

RonL

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