In a triangle ABC, the value of sin A + sin B | vedclass

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(B) In $\Delta ABC, A + B + C = 180^\circ.$$\sin A + \sin B + \sin C = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$Sin

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$4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

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(B) In $\Delta ABC, A + B + C = 180^\circ.$$\sin A + \sin B + \sin C = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + 2\sin \frac{C}{2} \cos \frac{C}{2}$Since $\frac{A+B}{2} = 90^\circ - \frac{C}{2}$,we have $\sin \frac{A+B}{2} = \cos \frac{C}{2}$.Also,$\sin \frac{C}{2} = \cos \frac{A+B}{2}$.Substituting these,we get:$= 2\cos \frac{C}{2} \cos \frac{A-B}{2} + 2\cos \frac{C}{2} \cos \frac{A+B}{2}$$= 2\cos \frac{C}{2} \left[ \cos \frac{A-B}{2} + \cos \frac{A+B}{2} \right]$Using the identity $\cos(x-y) + \cos(x+y) = 2\cos x \cos y$:$= 2\cos \frac{C}{2} \left( 2\cos \frac{A}{2} \cos \frac{B}{2} \right)$$= 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.

In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is

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