If tan (A + B) = p, tan (A – B) = q,

We know that,

tan 2A = tan (A + B + A – B)

And also,

$\tan (\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{x} \tan y}$

$\therefore \tan 2 \mathrm{~A}=\frac{\tan (\mathrm{A}+\mathrm{B})+\tan (\mathrm{A}-\mathrm{B})}{1-\tan (\mathrm{A}+\mathrm{B}) \tan (\mathrm{A}-\mathrm{B})}$

Substituting the values given in question,

$\Rightarrow \tan 2 A=\frac{\mathrm{p}+\mathrm{q}}{1-\mathrm{pq}}$

Hence, $\tan 2 \mathrm{~A}=\frac{\mathrm{p}+\mathrm{q}}{1-\mathrm{pq}}$