If

We have,

$\frac{\cos (A-B)}{\cos (A+B)}+\frac{\cos (C+D)}{\cos (C-D)}=0$

$\Rightarrow \frac{\cos (A-B) \cos (C-D)+\cos (C+D) \cos (A+B)}{\cos (A+B) \cos (C-D)}=0$

$\Rightarrow \cos (A-B) \cos (C-D)+\cos (C+D) \cos (A+B)=0$

$\Rightarrow \cos (A-B) \cos (C-D)=-\cos (C+D) \cos (A+B)$

$\Rightarrow[\cos A \cos B+\sin A \sin B][\cos C \cos D+\sin C \sin D]=$$-[\cos C \cos D-\sin C \sin D][\cos A \cos B-\sin A \sin B]$

Dividing both sides by $\cos A \cos B \cos C \cos D$ we get,

$\frac{[\cos A \cos B+\sin A \sin B][\cos C \cos D+\sin C \sin D]}{\cos A \cos B \cos C \cos D}=-\frac{[\cos C \cos D-\sin C \sin D][\cos A \cos B-\sin A \sin B]}{\cos A \cos B \cos C \cos D}$

$\Rightarrow \frac{[\cos A \cos B+\sin A \sin B]}{\cos A \cos B} \times \frac{[\cos C \cos D+\sin C \sin D]}{\cos C \cos D}=-\frac{[\cos C \cos D-\sin C \sin D]}{\cos C \cos D} \times \frac{[\sin C \cos A \cos B-\sin A \sin B]}{\cos A \cos B}$

$\Rightarrow[1+\tan A \tan B][1+\tan C \tan D]=[\tan C \tan D-1][1-\tan A \tan B]$

$\Rightarrow 1+\tan C \tan D+\tan A \tan B+\tan A \tan B \tan C \tan D=\tan C \tan D-\tan A \tan B \tan C \tan D+\tan A \tan B \tan D$$-1+\tan A \tan B$

$\Rightarrow 2 \tan A \tan B \tan C \tan D=-2$

$\Rightarrow \tan A \tan B \tan C \tan D=-1$

Hence proved.