B2. Let $x$ be such a real number for which $\cos \left(60^{\circ}-x\right) \neq 0$ and $\sin \left(120^{\circ}-x\right) \neq 0$. Without using a pocket calculator, calculate the exact value of the expression: $\frac{\sqrt{3}+4 \sin x \cos x}{\cos \left(60^{\circ}-x\right) \cdot \sin \left(120^{\circ}-x\right)}$.
B2. First, we use the addition theorems and simplify the expressions as much as possible $\cos \left(60^{\circ}-x\right)=$ $\cos 60^{\circ} \cdot \cos x+\sin 60^{\circ} \cdot \sin x=\frac{1}{2} \cdot \cos x+\frac{\sqrt{3}}{2} \cdot \sin x$ and $\sin \left(120^{\circ}-x\right)=\sin 120^{\circ} \cdot \cos x-\sin x \cdot \cos 120^{\circ}=$ $\frac{1}{2} \cdot \sin x+\frac{\sqrt{3}}{2} \cdot \cos x$. The product of the simplified expressions obtained is then $\frac{\sqrt{3}+4 \sin x \cos x}{4}$. The quotient $\frac{\sqrt{3}+4 \sin x \cos x}{\cos \left(60^{\circ}-x\right) \cdot \sin \left(120^{\circ}-x\right)}$ is equal to 4 after resolving the double fractions.

Application of the addition theorem $\cos \left(60^{\circ}-x\right)=\cos 60^{\circ} \cdot \cos x+\sin 60^{\circ} \cdot \sin x=\frac{1}{2} \cdot \cos x+\frac{\sqrt{3}}{2} \cdot \sin x 1+1$ point

Application of the addition theorem $\sin \left(120^{\circ}-x\right)=\sin 120^{\circ} \cdot \cos x-\sin x \cdot \cos 120^{\circ}=\frac{1}{2} \cdot \sin x+\frac{\sqrt{3}}{2} \cdot \cos x$ $1+1$ point

Calculation $\left(\frac{1}{2} \cdot \cos x+\frac{\sqrt{3}}{2} \cdot \sin x\right)\left(\frac{1}{2} \cdot \sin x+\frac{\sqrt{3}}{2} \cdot \cos x\right)=\frac{\sqrt{3}}{4} \sin ^{2} x+\frac{\sqrt{3}}{4} \cos ^{2} x+\sin x \cos x \ldots 1$ point Calculation $\frac{\sqrt{3}}{4}\left(\sin ^{2} x+\cos ^{2} x\right)+\sin x \cos x=\frac{\sqrt{3}}{4}+\sin x \cos x=\frac{\sqrt{3}+4 \sin x \cos x}{4} \ldots \ldots \ldots \ldots 1+1$ point