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If $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$, then $\tan A \tan B \tan C=$
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Verified Answer
The correct answer is:
$-\tan D$
$$
\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}
$$
By Componendo - Dividendo, we write
$$
\begin{aligned}
& \frac{\cos (A+B)+\cos (A-B)}{\cos (A+B)-\cos (A-B)}=\frac{\sin (C+D)+\sin (C-D)}{\sin (C+D)-\sin (C-D)} \\
& \frac{2 \cos A \cos B}{-2 \sin A \sin B}=\frac{2 \sin C \cos D}{2 \cos C \sin D} \\
& \therefore \frac{1}{-\tan A \tan B}=\frac{\tan C}{\tan D} \Rightarrow \tan A \tan B \tan C=-\tan D
\end{aligned}
$$
\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}
$$
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