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If $\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$, then $\tan A \cot B=$
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The correct answer is:
$-\cot C \cot D$
$\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$
$\therefore \frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}=\frac{\cos (C+D)+\cos (C-D)}{\cos (C+D)-\cos (C-D)}$
$\therefore \frac{2 \sin A \cos B}{2 \cos A \sin B}=\frac{2 \cos C \cos D}{-2 \sin C \sin D}$
$\quad \tan A \cot B=-\cot C \cot D$
$\therefore \frac{\sin (A+B)+\sin (A-B)}{\sin (A+B)-\sin (A-B)}=\frac{\cos (C+D)+\cos (C-D)}{\cos (C+D)-\cos (C-D)}$
$\therefore \frac{2 \sin A \cos B}{2 \cos A \sin B}=\frac{2 \cos C \cos D}{-2 \sin C \sin D}$
$\quad \tan A \cot B=-\cot C \cot D$
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