If frac{cos (A+B)}{cos (A-B)}=frac{sin (C+D)} | vedclass

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(D) Given: $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$Applying Componendo and Dividendo:$\frac{\cos (A+B) + \cos (A-B)}{\cos (A+B) -

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$-	an D$

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(D) Given: $\frac{\cos (A+B)}{\cos (A-B)}=\frac{\sin (C+D)}{\sin (C-D)}$Applying Componendo and Dividendo:$\frac{\cos (A+B) + \cos (A-B)}{\cos (A+B) - \cos (A-B)} = \frac{\sin (C+D) + \sin (C-D)}{\sin (C+D) - \sin (C-D)}$Using the identities $\cos(x+y) + \cos(x-y) = 2\cos x \cos y$,$\cos(x+y) - \cos(x-y) = -2\sin x \sin y$,$\sin(x+y) + \sin(x-y) = 2\sin x \cos y$,and $\sin(x+y) - \sin(x-y) = 2\cos x \sin y$:$\frac{2 \cos A \cos B}{-2 \sin A \sin B} = \frac{2 \sin C \cos D}{2 \cos C \sin D}$$-\cot A \cot B = 	an C \cot D$$-\frac{1}{	an A 	an B} = \frac{	an C}{	an D}$Therefore,$	an A 	an B 	an C = -	an D$

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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