If A lies in second quadrant 3tanA + 4 = 0,

It is given that $\frac{\pi}{2}

$3 \tan A+4=0$

$\Rightarrow \tan A=-\frac{4}{3}$

$\Rightarrow \cot A=-\frac{3}{4}$

Now,

$\sec A=\pm \sqrt{1+\tan ^{2} A}=\pm \sqrt{1+\frac{16}{9}}=\pm \sqrt{\frac{25}{9}}=\pm \frac{5}{3}$

$\therefore \sec A=-\frac{5}{3} \quad$ (A lies in 2 nd quadrant)

$\Rightarrow \cos A=-\frac{3}{5}$

Also,

$\sin A=\pm \sqrt{1-\cos ^{2} A}=\pm \sqrt{1-\frac{9}{25}}=\pm \sqrt{\frac{16}{25}}=\pm \frac{4}{5}$

$\therefore \sin A=\frac{4}{5}$

So,

$2 \cot A-5 \cos A+\sin A$

$=2 \times\left(-\frac{3}{4}\right)-5 \times\left(-\frac{3}{5}\right)+\frac{4}{5}$

$=-\frac{3}{2}+3+\frac{4}{5}$

$=\frac{-15+30+8}{10}$

$=\frac{23}{10}$

Hence, the correct answer is option B.