• # question_answer If A lies in the third quadrant and $3\,\tan A-4=0,$ then $5\,\sin 2A+3\,\sin A+4\,\cos A=$ [EAMCET 1994] A) 0 B) $\frac{-24}{5}$ C) $\frac{24}{5}$ D) $\frac{48}{5}$

$3\tan A-4=0\Rightarrow \tan A=\frac{4}{3}\Rightarrow \sin A=-\frac{4}{5},\cos A=-\frac{3}{5}$ $\therefore$ $5\sin 2A+3\sin A+4\cos A$ = $10\sin A\cos A+3\sin A+4\cos A$ = $10\,\left( \frac{12}{25} \right)-\frac{12}{5}-\frac{12}{5}=0$.