In a triangle ABC,if a^4+b^4+c^4=2b^2c^2+2a^2 | vedclass

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(A) Given,$a^4+b^4+c^4=2b^2c^2+2a^2b^2$.Rearranging the terms,we get $a^4+b^4+c^4-2b^2c^2-2a^2b^2=0$.Adding $2a^2c^2$ to both sides,we have $a^4+c^4+b

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$\frac{\pi}{4}$ or $\frac{3\pi}{4}$

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(A) Given,$a^4+b^4+c^4=2b^2c^2+2a^2b^2$.Rearranging the terms,we get $a^4+b^4+c^4-2b^2c^2-2a^2b^2=0$.Adding $2a^2c^2$ to both sides,we have $a^4+c^4+b^4-2a^2b^2-2b^2c^2+2a^2c^2 = 2a^2c^2$.This simplifies to $(a^2+c^2-b^2)^2 = 2a^2c^2$.Dividing by $4a^2c^2$,we get $\frac{(a^2+c^2-b^2)^2}{4a^2c^2} = \frac{2a^2c^2}{4a^2c^2} = \frac{1}{2}$.Since $\cos B = \frac{a^2+c^2-b^2}{2ac}$,we have $\cos^2 B = \frac{1}{2}$.Thus,$\cos B = \pm \frac{1}{\sqrt{2}}$.Therefore,$B = \frac{\pi}{4}$ or $B = \frac{3\pi}{4}$.

In a $\triangle ABC$,if $a^4+b^4+c^4=2b^2c^2+2a^2b^2$,then $B=$

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