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

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

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$ \ge 1$

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(B) Let $x = 	an \frac{A}{2}, y = 	an \frac{B}{2}, z = 	an \frac{C}{2}$.Since $A + B + C = \pi$,we have $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$.Using the identity $	an(\frac{A}{2} + \frac{B}{2} + \frac{C}{2}) = \frac{x+y+z-xyz}{1-(xy+yz+zx)} = 	an \frac{\pi}{2} = \infty$.This implies $1 - (xy + yz + zx) = 0$,so $xy + yz + zx = 1$.We know the inequality $(x-y)^2 + (y-z)^2 + (z-x)^2 \ge 0$.Expanding this,we get $2(x^2 + y^2 + z^2) - 2(xy + yz + zx) \ge 0$.$x^2 + y^2 + z^2 \ge xy + yz + zx$.Since $xy + yz + zx = 1$,we have $x^2 + y^2 + z^2 \ge 1$.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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