If $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$, then $\tan A \tan B \tan C=$

If $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$, then $\tan A \tan B \tan C=$
  1. 0
  2. $\tan \mathrm{D}$
  3. $\cot D$
  4. $-\tan D$

Solution

$\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$ By Componendo - Dividendo, we write $\begin{aligned} & \frac{\cos (A+B)+\cos (A-B)}{\cos (A+B)-\cos (A-B)}=\frac{\sin (C+D)+\sin (C-D)}{\sin (C+D)-\sin (C-D)} \\ & \frac{2 \cos A \cos B}{-2 \sin A \sin B}=\frac{2 \sin C \cos D}{2 \cos C \sin D} \\ & \therefore \frac{1}{-\tan A \tan B}=\frac{\tan C}{\tan D} \Rightarrow \tan A \tan B \tan C=-\tan D \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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