If $A, B, C$ are angles of a $\triangle A B C$, then $\tan 2 A+\tan 2 B+\tan 2 C=$

If $A, B, C$ are angles of a $\triangle A B C$, then $\tan 2 A+\tan 2 B+\tan 2 C=$
  1. $\tan 2 A \tan 3 B \tan 2 C$
  2. $\tan 2 A \tan 2 B \tan 2 C$
  3. $\tan A \tan B \tan C$
  4. $\tan 3 A \tan 2 B \tan 2 C$

Solution

$\begin{aligned} & \text { In } \Delta A B C, A+B+C=\pi \Rightarrow 2 A+2 B+2 C=2 \pi \\ \therefore & 2 A+2 B=2 \pi-2 C \Rightarrow \tan (2 A+2 B)=\tan (2 \pi-2 C)=-\tan 2 C \\ & \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \tan 2 B}=-\tan 2 C \\ \therefore & \tan 2 A+\tan 2 B=-\tan 2 C(1-\tan 2 A \tan 2 B) \\ \therefore & \tan 2 A+\tan 2 B+\tan 2 C=\tan 2 A \tan 2 B \tan 2 C \end{aligned}$

Asked in: MHT CET 2020 (15 Oct Shift 2)

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