# 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

$\begin{array}{1 1}(A)\;-\large\frac{53}{10}&(B)\;\large\frac{23}{10}\\(C)\;\large\frac{37}{10}&(D)\;\large\frac{7}{10}\end{array}$

$3\tan A+4=0$
$A$ lies in the second quadrant
$\tan A=-\large\frac{4}{3}$
$\sin A=\large\frac{4}{5}$
$\cos A=-\large\frac{3}{5}$
$2\cot A-5\cos A+\sin A$
$\Rightarrow 2\times \large\frac{-3}{4}$$+5\times \large\frac{3}{5}+\frac{4}{5} \Rightarrow -\large\frac{3}{2}$$+3+\large\frac{4}{5}$
$\Rightarrow \large\frac{-15+30+8}{10}$
$\Rightarrow \large\frac{23}{10}$
Hence (B) is the correct answer.