A) \[\frac{-53}{10}\]
B) \[\frac{-7}{10}\]
C) \[\frac{7}{10}\]
D) \[\frac{23}{10}\]
Correct Answer: D
Solution :
\[3\,\tan A+4=0\,\Rightarrow \,\tan A=-\frac{4}{3}\] \[\Rightarrow \,\,\sin A\,=\pm \,\frac{\tan A}{\sqrt{1+{{\tan }^{2}}A}}=\pm \frac{-4/3}{\sqrt{1+16/9}}=\frac{4}{5}\] \[(\because \,\,A\] is in 2nd quadrant) and \[\cos \,A=-\frac{3}{5}\]. Thus, \[2\cot A-5\cos A+\sin A\] \[=2\,\left( -\frac{3}{4} \right)-5\,\left( -\frac{3}{5} \right)+\frac{4}{5}=\frac{23}{10}\].
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