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(D) $A, B, C$ are angles of a triangle. Since $A + B + C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$. Using the sum-to-product f

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$5$

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(D) $A, B, C$ are angles of a triangle. Since $A + B + C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$. Using the sum-to-product formula: $\cos A + \cos B + \cos C = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right) + 1 - 2 \sin^2 \frac{C}{2}$ $= 2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right) + 1 - 2 \sin^2 \frac{C}{2}$ $= 1 + 2 \sin \frac{C}{2} \left[ \cos \left(\frac{A-B}{2}\right) - \sin \frac{C}{2} \right]$ $= 1 + 2 \sin \frac{C}{2} \left[ \cos \left(\frac{A-B}{2}\right) - \cos \left(\frac{A+B}{2}\right) \right]$ $= 1 + 2 \sin \frac{C}{2} \left[ 2 \sin \frac{A}{2} \sin \frac{B}{2} \right]$ $= 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$. Comparing this with $a + b \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,we get $a = 1$ and $b = 4$. Therefore,$a + b = 1 + 4 = 5$.

In a $\triangle ABC$,if $\cos A + \cos B + \cos C = a + b \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,then $(a + b)$ is equal to

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