Sum and Difference

**john doe** · November 10th 2008, 05:01 PM

$<br /> \tan(A + B)={-7/6},<br /> \tan(B)={5/3}<br />$

Find tan(A)

**skeeter** · November 10th 2008, 05:30 PM

use the sum identity ...

$\tan(a+b) = \frac{\tan{a} + \tan{b}}{1 - \tan{a}\tan{b}}$

sub in what you know, solve for $\tan{a}$

**john doe** · November 10th 2008, 05:56 PM

and how would i solve for tan(a), not sure where to start.

**skeeter** · November 11th 2008, 02:51 PM

$\tan(a+b) = \frac{\tan{a} + \tan{b}}{1 - \tan{a}\tan{b}}$

$\tan(a+b) (1 - \tan{a}\tan{b}) = \tan{a} + \tan{b}$

$\tan(a+b) - \tan(a+b) \tan{a} \tan{b} = \tan{a} + \tan{b}$

$\tan(a+b) - \tan{b} = \tan{a} - \tan(a+b) \tan{a} \tan{b}<br />$

$\tan(a+b) - \tan{b} = \tan{a}[1 - \tan(a+b) \tan{b}]$

$\frac{\tan(a+b) - \tan{b}}{1 - \tan(a+b) \tan{b}} = \tan{a}$

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