A) \[-53/10\]
B) \[-7/10\]
C) \[7/10\]
D) \[23/10\]
Correct Answer: D
Solution :
Key Idea: In second quadrant \[tan,\,\,cot,\,\,cos\] are negative and sin are positive. Given that, \[3\tan A+4=0\Rightarrow \tan A=-\frac{4}{3}\] Since, \[A\] lies in IInd quadrant \[\therefore \] \[\cot A=-\frac{3}{4},\,\,\cos A=-\frac{3}{5},\,\,\sin A=\frac{4}{5}\] \[\therefore \] \[2\cot A-5\cos A+\sin A\] \[=2\left( -\frac{3}{4} \right)-5\left( -\frac{3}{5} \right)+\frac{4}{5}\] \[=-\frac{3}{2}+3+\frac{4}{5}\] \[=\frac{-15+30+8}{10}=\frac{23}{10}\]You need to login to perform this action.
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