Statement-I : If $\mathrm{x} \mathrm{cosq} = \mathrm{y} \cos ({120}^0 + \mathrm{q}) = \mathrm{z}$ $\cos (\mathrm{q} + {240}^0)$ then $\mathrm{xy} + \mathrm{yz} + \mathrm{zx}=1$ 

Statement-II : $\cos \mathrm{a} + \cos ({120}^0 + \mathrm{a}) + \cos ({120}^0-\mathrm{a})=0$

Which of the above statements is correct

S-I:

$\begin{array}{rcl} \cos \mathrm{q}& =& \mathrm{y} \cos ({120}^0+\mathrm{q})=3\cos ({240}^0+\mathrm{q})=\mathrm{k}\\ \frac{\mathrm{k}}{\mathrm{x}}& =& \cos \mathrm{q}, \frac{\mathrm{k}}{\mathrm{y}}=\cos ({120}^0+\mathrm{q})\\ \frac{\mathrm{k}}{\mathrm{z}}& =& \cos ({240}^0+\mathrm{q})\\ \frac{\mathrm{k}}{\mathrm{x}}+\frac{\mathrm{k}}{\mathrm{y}}+\frac{\mathrm{k}}{\mathrm{z}}& =& \cos \mathrm{q}+\cos ({180}^0-(60-\mathrm{q}\left)\right)+\cos ({180}^0+{60}^0+\mathrm{q})\\ & =& \cos \mathrm{q}-\cos (60-\mathrm{q})-\cos ({60}^0+\mathrm{q})\\ & =& \cos \mathrm{q}-2\frac{1}{2}\cos \mathrm{q}\\ & =& 0\\ \frac{\mathrm{k}(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})}{\mathrm{xyz}}& =& 0\end{array}$

         $\mathrm{xy}+\mathrm{y}+\mathrm{zx}=0$

S-II:

$\begin{array}{rcl} & & \cos \mathrm{a}+\cos ({120}^0+\mathrm{a})+\cos ({120}^0-\mathrm{a})\\ & =& \cos \mathrm{a}+2\cos {120}^0 \cos \mathrm{a}\\ & =& \cos \mathrm{a}+2\left(\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right) \cos \mathrm{a}\\ & =& \cos \mathrm{a}-\cos \mathrm{a}\\ & =& 0\end{array}$