if tan(a-b)=1/4, tan(a+b)=1/3 then tan2a=?

$$\\atan(\frac{1}{4})=A-B\qquad (1)\\\\ atan(\frac{1}{3})=A+B\qquad(2)\\\\ (1)+(2)\\\\ atan(\frac{1}{4})+atan(\frac{1}{3})=2A\\\\ tan[atan(\frac{1}{4})+atan(\frac{1}{3})]=tan[2A]\\\\$$

$$$Now I am going to use the identity$\\\\ \boxed{tan(\theta_1+\theta_2)=\frac{tan(\theta_1)+tan(\theta_2)}{1-tan(\theta_1)tan(\theta_2)}}\\\\\\ tan(2A)\\\\ =tan[atan(\frac{1}{4})+atan(\frac{1}{3})]\\\\ =\frac{\frac{1}{4}+\frac{1}{3}}{1-\frac{1}{4}\times\frac{1}{3}}\\\\ =\frac{\frac{7}{12}}{1-\frac{1}{12}}\\\\ =\frac{7}{12}\div \frac{11}{12}\\\\ =\frac{7}{12}\times \frac{12}{11}\\\\ =\frac{7}{11}\\\\ $This answer is exact$$ $

Using the tangent inverse

tan^-1(1/4) = (a - b) = about 14.04°  ......  and tan^-1(1/3) = (a +b) = about 18.43°

So

(a -b) + (a+b) = (14.04 + 18.43)° →  

2a = about 32.47°

So

tan(2a) = tan(32.47°) = about .636363= 7/11

Thanks Chris    

I did mine "on the fly"....Melody's approach is actually "better"