If A, B, C are the angles of a triangle,then | vedclass

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(B) Given the expression: $\frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}-\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1}$ Using $\sin A + \sin B = 2 \sin

Vedclass · Accepted Answer

$-2 \cot \frac{B}{2}$

Vedclass · Answer

(B) Given the expression: $\frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}-\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1}$ Using $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2}$,the numerator becomes $2 \cos \frac{C}{2} (\cos \frac{A-B}{2} + \sin \frac{C}{2})$. Since $A+B+C = \pi$,$\sin \frac{C}{2} = \cos \frac{A+B}{2}$. Numerator $= 2 \cos \frac{C}{2} (\cos \frac{A-B}{2} + \cos \frac{A+B}{2}) = 2 \cos \frac{C}{2} (2 \cos \frac{A}{2} \cos \frac{B}{2}) = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$. For the denominator: $\sin ^2 \frac{A}{2} - \sin ^2 \frac{B}{2} + \sin ^2 \frac{C}{2} - 1 = \sin ^2 \frac{A}{2} - \sin ^2 \frac{B}{2} - \cos ^2 \frac{C}{2} = \sin(\frac{A-B}{2}) \sin(\frac{A+B}{2}) - \cos^2 \frac{C}{2}$. Using $\sin(\frac{A+B}{2}) = \cos \frac{C}{2}$,denominator $= \cos \frac{C}{2} (\sin \frac{A-B}{2} - \cos \frac{C}{2}) = \cos \frac{C}{2} (\sin \frac{A-B}{2} - \sin \frac{A+B}{2}) = \cos \frac{C}{2} (-2 \cos \frac{A}{2} \sin \frac{B}{2})$. Dividing numerator by denominator: $\frac{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{-2 \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}} = -2 \cot \frac{B}{2}$.

If $A, B, C$ are the angles of a triangle,then $\frac{\sin A+\sin B+\sin C}{\sin ^2 \frac{A}{2}-\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}-1} =$

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