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

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

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$-\cot C \cot D$

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

If $\frac{\sin (A+B)}{\sin (A-B)}=\frac{\cos (C+D)}{\cos (C-D)}$,then $\tan A \cot B=$

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