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$.Thus,$1 + 	an B + 	an

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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$.Thus,$1 + 	an B + 	an C = 	an B 	an C = K$.$	an B + 	an C = K - 1$.We know that $	an B$ and $	an C$ are roots of the quadratic equation $x^2 - (	an B + 	an C)x + 	an B 	an C = 0$.Substituting the values,$x^2 - (K - 1)x + K = 0$.For $	an B$ and $	an C$ to be real,the discriminant $D \geqslant 0$.$D = (K - 1)^2 - 4K \geqslant 0$.$K^2 - 2K + 1 - 4K \geqslant 0$.$K^2 - 6K + 1 \geqslant 0$.

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

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