જો A + B + C = 180^circ હોય,તો frac{sin 2A + | vedclass

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(B) આપેલ છે કે $A + B + C = 180^\circ.$અંશ $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 32\sin \frac{A}{2}\cos \frac{A}{2}\sin \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ છે કે $A + B + C = 180^\circ.$અંશ $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 32\sin \frac{A}{2}\cos \frac{A}{2}\sin \frac{B}{2}\cos \frac{B}{2}\sin \frac{C}{2}\cos \frac{C}{2}.$છેદ $(D^r)$ = $\cos A + \cos B + \cos C - 1 = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$તેથી,$\frac{N^r}{D^r} = \frac{32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} = 8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$

જો $A + B + C = 180^\circ$ હોય,તો $\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = $

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