Tính A = sin150° + tan135° + cot45°; B = 2cos30°
Giải thích
Ta có A = sin150° + tan135° + cot45° = 12+(−1)+1= 12.
B = 2cos30° – 3tan150° + cot135° = 2.32−3.−33+(−1)= 23−1.
Vậy A = 12 ; B = 23−1 .
Ta có A = sin150° + tan135° + cot45° = 12+(−1)+1= 12.
B = 2cos30° – 3tan150° + cot135° = 2.32−3.−33+(−1)= 23−1.
Vậy A = 12 ; B = 23−1 .