If A + B + C = π, prove that

= tan 2A + tan 2B + tan 2C

Since $A+B+C=\pi$

$A+B=\pi-C$

$2 A+2 B=2 \pi-2 C$

$\operatorname{Tan}(2 A+2 B)=\tan (2 \pi-2 C)$

Since $\tan (2 \pi-C)=-\tan C$

$\operatorname{Tan}(2 A+2 B)=-\tan 2 C$

Now using formula,

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$\frac{\tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}}{1-\tan 2 \mathrm{~A} \tan 2 \mathrm{~B}}=-\tan 2 \mathrm{C}$

Tan $2 A+\tan 2 B=-\tan 2 C+\tan 2 C \tan 2 B \tan 2 A$

Tan $2 A+\tan 2 B+\tan 2 C=\tan 2 A \tan 2 B \tan 2 C$

= R.H.S