Lời giải:
Ta có: \(\cos a=\frac{-2}{3}\Rightarrow \sin ^2a=1-\cos ^2a=\frac{5}{9}\)
\(B=\frac{\cot a+3\tan a}{2\cot a+3\tan a}=\frac{\frac{1}{\tan a}+3\tan a}{\frac{2}{\tan a}+3\tan a}=\frac{1+3\tan ^2a}{2+3\tan ^2a}\)
Lại có: \(\tan ^2a=\frac{\sin ^2a}{\cos ^2a}=\frac{\frac{5}{9}}{\frac{4}{9}}=\frac{5}{4}\)
Do đó \(B=\frac{1+3.\frac{5}{4}}{2+3.\frac{5}{4}}=\frac{19}{23}\)