In a triangle ABC,if frac{a}{tan A}=frac{b}{t | vedclass

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(B) Given $\frac{a}{	an A} = \frac{b}{	an B} = \frac{c}{	an C}$.Since $	an A = \frac{\sin A}{\cos A}$,we have $\frac{a \cos A}{\sin A} = \frac{b \

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$\frac{3}{4}$

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(B) Given $\frac{a}{	an A} = \frac{b}{	an B} = \frac{c}{	an C}$.Since $	an A = \frac{\sin A}{\cos A}$,we have $\frac{a \cos A}{\sin A} = \frac{b \cos B}{\sin B} = \frac{c \cos C}{\sin C}$.By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.Substituting this into the given equation,we get $2R \cos A = 2R \cos B = 2R \cos C$.This implies $\cos A = \cos B = \cos C$.Since $A, B, C$ are angles of a triangle,$A = B = C = 60^{\circ}$.Therefore,$\cos^2 60^{\circ} + \cos^2 60^{\circ} + \cos^2 60^{\circ} = (\frac{1}{2})^2 + (\frac{1}{2})^2 + (\frac{1}{2})^2 = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.

In a triangle $ABC$,if $\frac{a}{\tan A}=\frac{b}{\tan B}=\frac{c}{\tan C}$,then $\cos^2 A + \cos^2 B + \cos^2 C =$

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