In a triangle ABC,if A = frac{pi}{4} and tan | vedclass

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(A) In a $	riangle ABC$,we know that $	an A + 	an B + 	an C = 	an A 	an B 	an C$.Since $A = \frac{\pi}{4}$,$	an A = 1$.Substituting this,we ge

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$K^2 - 6K + 1 \geqslant 0$

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(A) In a $	riangle ABC$,we know that $	an A + 	an B + 	an C = 	an A 	an B 	an C$.Since $A = \frac{\pi}{4}$,$	an A = 1$.Substituting this,we get $1 + 	an B + 	an C = 	an B 	an C$.Given $	an B 	an C = K$,we have $	an B + 	an C = K - 1$.Since $	an B$ and $	an C$ are roots of the quadratic equation $x^2 - (	an B + 	an C)x + 	an B 	an C = 0$,we have $x^2 - (K - 1)x + K = 0$.For $	an B$ and $	an C$ to be real,the discriminant $D$ must be $\geqslant 0$.$D = (K - 1)^2 - 4K \geqslant 0$.$K^2 - 2K + 1 - 4K \geqslant 0$.$K^2 - 6K + 1 \geqslant 0$.

In a triangle $ABC$,if $A = \frac{\pi}{4}$ and $\tan B \tan C = K$,then $K$ must satisfy:

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