If cos (A + B) sin (C − D) = cos (A − B) sin (C + D),

cos (A + B) sin (C − D) = cos (A − B) sin (C + D)

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

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

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

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

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

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

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

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

$\Rightarrow \tan A \tan B \tan C+\tan D=0$

Hence proved.