# If $\tan(A+B)=p$ and $\tan(A-B)=q$ then show that $\tan 2A=\large\frac{p+q}{1-pq}$

Toolbox:
• $\tan (a+b)=\large\frac{\tan a+\tan b}{1-\tan a.\tan b}$
Given :
$\tan(A+B)=p$ and $\tan (A-B)=q$
L.H.S
$\tan 2A=\tan(A+B+A-B)$
$\tan[(A+B)+(A-B)]=\large\frac{\tan(A+B)+\tan (A-B)}{1-\tan (A+B)\tan (A-B)}=\frac{p+q}{1-pq}$
Hence proved.