[tex]\it x\in(0,\ \dfrac{\pi}{2}),\ \ cos x=\dfrac{1}{2}\\ \\ \\ sinx=\sqrt{1-cos^2x}=\sqrt{1-\dfrac{1}{4}}=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt3}{2}\\ \\ \\ tgx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{2}}=\dfrac{\sqrt3}{2}:\dfrac{1}{2}=\dfrac{\sqrt3}{2}\cdot\dfrac{2}{1}=\sqrt3[/tex]
