If A lies in the third quadrant and 3 tan A - 4 = 0, then 5 sin 2A + 3 sin A + 4 cos A =

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Solution

The correct option is A
$3 t a n A - 4 = 0 \Rightarrow t a n A = \frac{4}{3} \Rightarrow s i n A = - \frac{4}{5}, c o s A = - \frac{3}{5}$
$∴$ 5 sin 2A + 3 sin A + 4 cos A
=10 sin A cos A + 3 sin A + 4 cos A
$= 10 (\frac{12}{25}) - \frac{12}{5} - \frac{12}{5} = 0.$

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