proving trig identities

**Tweety** · November 11th 2009, 01:57 PM

$cot(A+B) = \frac{cotAcotB-1}{cotA+cotB}$

so far I have done;
$\frac{ \frac{1}{tanA} \times \frac{1}{tanB} - 1}{ \frac{1}{tanA} + \frac{1}{tanB}}$

which after simplifying I get; $\frac{1-tanAtanB}{tanA + tanB}$

I am stuck here, if anyone could direct me from here, thanks.

**stapel** · November 11th 2009, 02:09 PM

How does your result relate to the tangent identity they gave you?

**skeeter** · November 11th 2009, 02:10 PM

Originally Posted by Tweety

$cot(A+B) = \frac{cotAcotB-1}{cotA+cotB}$

so far I have done;
$\frac{ \frac{1}{tanA} \times \frac{1}{tanB} - 1}{ \frac{1}{tanA} + \frac{1}{tanB}}$

which after simplifying I get; $\frac{1-tanAtanB}{tanA + tanB}$

I am stuck here, if anyone could direct me from here, thanks.

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

**Tweety** · November 12th 2009, 12:55 AM

Originally Posted by skeeter

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

thanks,

But I dont understand how you get $\frac{1}{\tan(A+B)}$

from $\frac{1-\tan{A}\tan{B}}{\tan{A} + \tan{B}}$

Is there some rule I am suppose to know? I understand that the denominator was factored, but what happened to $1-tanA tanB$ ? How does it become 1?

**stapel** · November 12th 2009, 03:54 AM

Originally Posted by Tweety

Is there some rule I am suppose to know?

Yes: the tangent identity mentioned earlier. Check your chart or table of identities....

**Tweety** · November 12th 2009, 04:14 AM

Originally Posted by stapel

Yes: the tangent identity mentioned earlier. Check your chart or table of identities....

Oh I see what you mean, its this identity $tan(A + B) = \frac{tanA + tanB}{1-tanAtanB}$ but 'flipped' around?

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