If A lies in the second quadrant and $3\tan A+4=0,$ then the value of $2\cot A-5\cos A+\sin A$ is equal to

Using right angled triangle, we have

$$\begin{gathered} \tan A=-\frac{4}{3}\Rightarrow cotA=-\frac{3}{4}, \cos A=-\frac{3}{5} and \sin A=\frac{4}{5} \\ \therefore 2\cot A-5\cos A+\sin A=-\frac{3}{2}+3+\frac{4}{5}=\frac{23}{10} \end{gathered}$$