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

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(B) Given $A + B + C = 180^\circ$. Numerator: $\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 \f

Vedclass · Accepted Answer

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

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(B) Given $A + B + C = 180^\circ$. Numerator: $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C = 4(2 \sin \frac{A}{2} \cos \frac{A}{2})(2 \sin \frac{B}{2} \cos \frac{B}{2})(2 \sin \frac{C}{2} \cos \frac{C}{2}) = 32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}$. Denominator: $\cos A + \cos B + \cos C - 1 = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. Dividing the numerator by the denominator: $\frac{32 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}} = 8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

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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