a.

\(P=cos120^0+cos120^0+cos120^0=-\dfrac{3}{2}\)

b.

\(A=\dfrac{\dfrac{sinx}{cosx}-\dfrac{cosx}{cosx}}{\dfrac{sinx}{cosx}+\dfrac{cosx}{cosx}}=\dfrac{tanx-1}{tanx+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)

c.

\(A=\dfrac{cos\left(720+30\right)+sin\left(360+60\right)}{sin\left(-360+30\right)-cos\left(-360-30\right)}=\dfrac{cos30+sin60}{sin30-cos30}=-3-\sqrt{3}\)

Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)

\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)

\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:

a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)

\(=\cos150^o+\sin30^o+\tan60^o\)

\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)

\(=\frac{\sqrt{3}+1}{2}\)

b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)

\(=\sin90^o+\cos30^o+\cos0^o\)

\(=1+\frac{\sqrt{3}}{2}\)

\(=\frac{2+\sqrt{3}}{2}\)

Câu 1: 

\(1+\cot^2a=\dfrac{1}{\sin^2a}\)

nên \(\dfrac{1}{\sin^2a}=1+5^2=26\)

\(\Leftrightarrow\sin^2a=\dfrac{1}{26}\)

\(\Leftrightarrow\sin a=\dfrac{\sqrt{26}}{26}\)

\(\cos a=\sqrt{1-\dfrac{1}{26}}=\dfrac{5\sqrt{26}}{26}\)

\(A=\dfrac{\sin a+\cos a}{\sin a-\cos a}=\left(\dfrac{\sqrt{26}+5\sqrt{26}}{26}\right):\left(\dfrac{\sqrt{26}-5\sqrt{26}}{26}\right)\)

\(=\dfrac{6\sqrt{26}}{-4\sqrt{26}}=\dfrac{-3}{2}\)

Áp dụng hệ quả của định lí cô sin trong tam giác ta có: 

  c o s ⁡ A = ( b 2 + c 2 - a 2 ) / 2 b c = ( 5 2 + 4 2 - 6 2 ) / 2 . 5 . 4 = 1 / 8 = 0 , 125 .

Chọn A.