In triangle ABC,if cos A cdot cos B cdot cos | vedclass

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(C) Given: $\cos A \cos B \cos C = \frac{1}{5}$.In $	riangle ABC$,$A+B+C = \pi$,so $A+B = \pi - C$.Taking cosine on both sides: $\cos(A+B) = \cos(\pi

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$6$

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(C) Given: $\cos A \cos B \cos C = \frac{1}{5}$.In $	riangle ABC$,$A+B+C = \pi$,so $A+B = \pi - C$.Taking cosine on both sides: $\cos(A+B) = \cos(\pi - C) = -\cos C$.$\cos A \cos B - \sin A \sin B = -\cos C$.$\sin A \sin B = \cos A \cos B + \cos C$.Dividing by $\cos A \cos B$,we get $	an A 	an B = 1 + \frac{\cos C}{\cos A \cos B}$.Similarly,$	an B 	an C = 1 + \frac{\cos A}{\cos B \cos C}$ and $	an C 	an A = 1 + \frac{\cos B}{\cos C \cos A}$.Adding these three equations:$\sum 	an A 	an B = 3 + \frac{\cos^2 C + \cos^2 A + \cos^2 B}{\cos A \cos B \cos C}$.Using the identity $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C$ for $	riangle ABC$:$\sum 	an A 	an B = 3 + \frac{1 - 2 \cos A \cos B \cos C}{\cos A \cos B \cos C} = 3 + \frac{1}{\cos A \cos B \cos C} - 2$.Substituting $\cos A \cos B \cos C = \frac{1}{5}$:$\sum 	an A 	an B = 3 + 5 - 2 = 6$.

In $\triangle ABC$,if $\cos A \cdot \cos B \cdot \cos C = \frac{1}{5}$,then $\tan A \tan B + \tan B \tan C + \tan C \tan A = $

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