Ta có: `sin^2alpha+cos^2alpha=1`

`=> cos^2alpha=1-sin^2alpha=1-(-12/13)^2=25/169`

`=> cos alpha=+-\sqrt{25/169}=+-5/13`

Mà `(3pi)/2 < alpha <2pi` nên `cos alpha>0`

Hay `cos alpha=5/13`

`=> tan alpha=(sin alpha)/(cos alpha)=(-12/13)/(5/13)=-12/5`

Ta có: `tan (alpha+pi/3)=(tan alpha+tan(pi/3))/(1-tan alpha.tan(pi/3))`

`=(-12/5+\sqrt{3})/(1-(-12/5).\sqrt{3})\ne (48-3)/11`

`=>` Khẳng định trên là sai