If A + B + C = pi,then tan^2 frac{A}{2} + tan | vedclass

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(B) Let $x = 	an \frac{A}{2}$,$y = 	an \frac{B}{2}$,and $z = 	an \frac{C}{2}$.Since $A + B + C = \pi$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}

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$\ge 1$

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(B) Let $x = 	an \frac{A}{2}$,$y = 	an \frac{B}{2}$,and $z = 	an \frac{C}{2}$.Since $A + B + C = \pi$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$.Using the identity for $	an(X+Y+Z)$,we know that if $X+Y+Z = \frac{\pi}{2}$,then $xy + yz + zx = 1$.We know the algebraic inequality $(x-y)^2 + (y-z)^2 + (z-x)^2 \ge 0$.Expanding this,we get $2(x^2 + y^2 + z^2) - 2(xy + yz + zx) \ge 0$.Dividing by $2$,we get $x^2 + y^2 + z^2 \ge xy + yz + zx$.Substituting the value $xy + yz + zx = 1$,we get $x^2 + y^2 + z^2 \ge 1$.Thus,$	an^2 \frac{A}{2} + 	an^2 \frac{B}{2} + 	an^2 \frac{C}{2} \ge 1$.

If $A + B + C = \pi$,then $\tan^2 \frac{A}{2} + \tan^2 \frac{B}{2} + \tan^2 \frac{C}{2}$ is always

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