In a triangle ABC,if a^2-b^2-c^2=bc(lambda^2- | vedclass

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(C) Given the equation: $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$ Rearranging the terms,we get: $b^2+c^2-a^2 = -bc(\lambda^2-2\lambda-1)$ Dividing both s

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$-1 \leq \lambda \leq 3$

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(C) Given the equation: $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$ Rearranging the terms,we get: $b^2+c^2-a^2 = -bc(\lambda^2-2\lambda-1)$ Dividing both sides by $2bc$,we have: $\frac{b^2+c^2-a^2}{2bc} = -\frac{1}{2}(\lambda^2-2\lambda-1)$ Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so: $\cos A = -\frac{1}{2}(\lambda^2-2\lambda-1)$ Since $-1 \leq \cos A \leq 1$,we have: $-1 \leq -\frac{1}{2}(\lambda^2-2\lambda-1) \leq 1$ Multiplying by $-2$ (and reversing the inequalities): $-2 \leq \lambda^2-2\lambda-1 \leq 2$ Adding $1$ to all parts: $-1 \leq \lambda^2-2\lambda \leq 3$ Completing the square: $-1 \leq (\lambda-1)^2-1 \leq 3$ Adding $1$ to all parts: $0 \leq (\lambda-1)^2 \leq 4$ Taking the square root: $0 \leq |\lambda-1| \leq 2$ This implies $-2 \leq \lambda-1 \leq 2$,which simplifies to: $-1 \leq \lambda \leq 3$

In a triangle $ABC$,if $a^2-b^2-c^2=bc(\lambda^2-2\lambda-1)$,then

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