# Sum and Difference

• November 10th 2008, 05:01 PM
john doe
Sum and Difference
$
\tan(A + B)={-7/6},
\tan(B)={5/3}
$

Find tan(A)
• November 10th 2008, 05:30 PM
skeeter
use the sum identity ...

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

sub in what you know, solve for $\tan{a}$
• November 10th 2008, 05:56 PM
john doe
and how would i solve for tan(a), not sure where to start.
• November 11th 2008, 02:51 PM
skeeter
$\tan(a+b) = \frac{\tan{a} + \tan{b}}{1 - \tan{a}\tan{b}}$

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

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

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

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

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