If 4 times the area of a triangle ABC is c^2- | vedclass

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(D) Given,$4 \Delta = c^2 - (a - b)^2$. Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get: $4 \Delta = (c - (a - b))(c + (a - b)) = (c - a + b)(c

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(D) Given,$4 \Delta = c^2 - (a - b)^2$. Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get: $4 \Delta = (c - (a - b))(c + (a - b)) = (c - a + b)(c + a - b)$. Dividing by $4$,we have: $\Delta = \left(\frac{b + c - a}{2}ight) \left(\frac{a + c - b}{2}ight) = (s - a)(s - b)$,where $s$ is the semi-perimeter. From Heron's formula,$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$. Thus,$\sqrt{s(s - a)(s - b)(s - c)} = (s - a)(s - b)$. Squaring both sides: $s(s - a)(s - b)(s - c) = (s - a)^2(s - b)^2$. $\frac{s(s - c)}{(s - a)(s - b)} = 1$. We know that $	an^2\left(\frac{C}{2}ight) = \frac{(s - a)(s - b)}{s(s - c)}$. Therefore,$	an^2\left(\frac{C}{2}ight) = 1$,which implies $	an\left(\frac{C}{2}ight) = 1$. $\frac{C}{2} = \frac{\pi}{4} \Rightarrow C = \frac{\pi}{2}$. Hence,$\sin C = \sin\left(\frac{\pi}{2}ight) = 1$.

If $4$ times the area of a $\triangle ABC$ is $c^2-(a-b)^2$,then $\sin C$ is equal to:

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