If ABC is not a right-angled triangle and sin | vedclass

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(B) Given,$A+B+C = \pi$. We know that $	an A + 	an B + 	an C = 	an A 	an B 	an C$. The given equation is $\sin \left(\frac{\pi}{4}-Aight) \sin

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$	an A + 	an B + 	an C$

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(B) Given,$A+B+C = \pi$. We know that $	an A + 	an B + 	an C = 	an A 	an B 	an C$. The given equation is $\sin \left(\frac{\pi}{4}-Aight) \sin \left(\frac{\pi}{4}-Bight) \sin \left(\frac{\pi}{4}-Cight) = -\frac{1}{2 \sqrt{2}}$. Using $\sin \left(\frac{\pi}{4}-	hetaight) = \frac{1}{\sqrt{2}}(\cos 	heta - \sin 	heta) = \frac{\cos 	heta}{\sqrt{2}}(1 - 	an 	heta)$,we get: $\frac{\cos A \cos B \cos C}{2 \sqrt{2}} (1 - 	an A)(1 - 	an B)(1 - 	an C) = -\frac{1}{2 \sqrt{2}}$. $(1 - 	an A)(1 - 	an B)(1 - 	an C) = -\frac{1}{\cos A \cos B \cos C}$. Expanding the left side: $1 - (	an A + 	an B + 	an C) + (	an A 	an B + 	an B 	an C + 	an C 	an A) - 	an A 	an B 	an C = -\frac{1}{\cos A \cos B \cos C}$. Since $	an A + 	an B + 	an C = 	an A 	an B 	an C$,the terms cancel out: $1 + (	an A 	an B + 	an B 	an C + 	an C 	an A) - 2(	an A + 	an B + 	an C) = -\frac{1}{\cos A \cos B \cos C}$. For $A+B+C = \pi$,$\cos(A+B+C) = \cos A \cos B \cos C (1 - (	an A 	an B + 	an B 	an C + 	an C 	an A)) = -1$. Thus,$1 - (	an A 	an B + 	an B 	an C + 	an C 	an A) = -\frac{1}{\cos A \cos B \cos C}$. Equating the two expressions for $-\frac{1}{\cos A \cos B \cos C}$: $1 - (	an A 	an B + 	an B 	an C + 	an C 	an A) = 1 - (	an A + 	an B + 	an C) + (	an A 	an B + 	an B 	an C + 	an C 	an A) - 	an A 	an B 	an C$. Since $	an A + 	an B + 	an C = 	an A 	an B 	an C$,we get: $1 - (	an A 	an B + 	an B 	an C + 	an C 	an A) = 1 - 2(	an A + 	an B + 	an C) + (	an A 	an B + 	an B 	an C + 	an C 	an A)$. $2(	an A 	an B + 	an B 	an C + 	an C 	an A) = 2(	an A + 	an B + 	an C)$. Therefore,$	an A 	an B + 	an B 	an C + 	an C 	an A = 	an A + 	an B + 	an C$.

If $ABC$ is not a right-angled triangle and $\sin \left(\frac{\pi}{4}-A\right) \sin \left(\frac{\pi}{4}-B\right) = -\frac{1}{2 \sqrt{2}} \operatorname{cosec}\left(\frac{\pi}{4}-C\right)$,then $\tan A \tan B + \tan B \tan C + \tan C \tan A = $

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