In a triangle ABC with usual notations,if tan | vedclass

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(A) Given that $	an A, 	an B, 	an C$ are in $H.P.$$\frac{2}{	an B} = \frac{1}{	an A} + \frac{1}{	an C}$$\frac{2 \cos B}{\sin B} = \frac{\cos A}{

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(A) Given that $	an A, 	an B, 	an C$ are in $H.P.$$\frac{2}{	an B} = \frac{1}{	an A} + \frac{1}{	an C}$$\frac{2 \cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C}$Using the sine rule $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.Also,$\cos B = \frac{a^{2}+c^{2}-b^{2}}{2ac}, \cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}, \cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}$.Substituting these values:$2 \left( \frac{a^{2}+c^{2}-b^{2}}{2ac} ight) \cdot \frac{2R}{b} = \frac{b^{2}+c^{2}-a^{2}}{2bc} \cdot \frac{2R}{a} + \frac{a^{2}+b^{2}-c^{2}}{2ab} \cdot \frac{2R}{c}$Multiplying both sides by $\frac{abc}{2R}$:$2(a^{2}+c^{2}-b^{2}) = (b^{2}+c^{2}-a^{2}) + (a^{2}+b^{2}-c^{2})$$2a^{2} + 2c^{2} - 2b^{2} = 2b^{2}$$2a^{2} + 2c^{2} = 4b^{2}$$a^{2} + c^{2} = 2b^{2}$Thus,$a^{2}, b^{2}, c^{2}$ are in $A.P.$

In a triangle $ABC$ with usual notations,if $\tan A, \tan B, \tan C$ are in $H.P.$,then $a^{2}, b^{2}, c^{2}$ are in

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