If \( \tan(A+B)=p \) and \( \tan(A-B)=q \), then find \( \tan2A \)
Solution:
Since,
\[
2A=(A+B)+(A-B)
\]
Using identity,
\[
\tan(x+y)
=
\frac{\tan x+\tan y}{1-\tan x\tan y}
\]
\[
\tan2A
=
\tan[(A+B)+(A-B)]
\]
\[
=
\frac{\tan(A+B)+\tan(A-B)}
{1-\tan(A+B)\tan(A-B)}
\]
Given,
\[
\tan(A+B)=p
\]
and
\[
\tan(A-B)=q
\]
Substituting,
\[
\tan2A
=
\frac{p+q}{1-pq}
\]
\[
\boxed{\tan2A=\frac{p+q}{1-pq}}
\]