ta có cos60=1/2
sin 60=\(\frac{\sqrt{3}}{2}\)
tan 30=\(\frac{\sqrt{3}}{3}\)
ta thay vào biểu thức trên
=> \(\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=2\)
\(\frac{cos60^o}{1+sin60^o}+\frac{1}{tan30^o}=\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}+\frac{1}{\frac{\sqrt{3}}{3}}=\frac{1}{2}.\frac{2}{\sqrt{3}+2}+\sqrt{3}=\frac{1}{\sqrt{3}+2}+\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}+\sqrt{3}=2-\sqrt{3}+\sqrt{3}=2\)