Tính A = sin150° + tan135° + cot45°; B = 2cos30°

Ta có A = sin150° + tan135° + cot45° = $\frac{1}{2}+(-1)+1$ = $\frac{1}{2}$ .

B = 2cos30° – 3tan150° + cot135° = $2.\frac{\sqrt{3}}{2}-3.\left(-\frac{\sqrt{3}}{3}\right)+(-1)$ = $2\sqrt{3}-1$ .

Vậy A = $\frac{1}{2}$  ; B = $2\sqrt{3}-1$ .