If A + B + C = 180^circ, then frac{sin 2A + s | vedclass

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(B) Given $A + B + C = 180^\circ.$Numerator $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\si

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$8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

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(B) Given $A + B + C = 180^\circ.$Numerator $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\sin \frac{B}{2}\cos \frac{B}{2})(2\sin \frac{C}{2}\cos \frac{C}{2}) = 32\sin \frac{A}{2}\cos \frac{A}{2}\sin \frac{B}{2}\cos \frac{B}{2}\sin \frac{C}{2}\cos \frac{C}{2}.$Denominator $(D^r)$ = $\cos A + \cos B + \cos C - 1 = -4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$Wait,using the identity $\cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},$ we have $D^r = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$Now,$N^r = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\sin \frac{B}{2}\cos \frac{B}{2})(2\sin \frac{C}{2}\cos \frac{C}{2}) = 32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$Therefore,$\frac{N^r}{D^r} = \frac{32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} = 8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$

If $A + B + C = 180^\circ,$ then $\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = $

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