A) \[\frac{p+q}{p-q}\]
B) \[\frac{p-q}{1+pq}\]
C) \[\frac{p+q}{1-pq}\]
D) \[\frac{1+pq}{1-p}\]
Correct Answer: C
Solution :
\[2A=(A+B)+(A-B)\] \[\Rightarrow \]\[\tan 2A=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B)\tan (A-B)}=\frac{p+q}{1-pq}\].
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