MCQ
If $A$ lies in the third quadrant and $3\,\tan A - 4 = 0,$ then $5\,\sin 2A + 3\,\sin A + 4\,\cos A = $
- ✓$0$
- B$\frac{{ - 24}}{5}$
- C$\frac{{24}}{5}$
- D$\frac{{48}}{5}$
✓
Answer
Correct option: A.
$0$
a
(a) $3\tan A - 4 = 0 \Rightarrow \tan A = \frac{4}{3} $
(a) $3\tan A - 4 = 0 \Rightarrow \tan A = \frac{4}{3} $
$\Rightarrow \sin A = - \frac{4}{5},\cos A = - \frac{3}{5}$
$\therefore $ $5\sin 2A + 3\sin A + 4\cos A$
$= 10\sin A\cos A + 3\sin A + 4\cos A$
$= 10\,\left( {\frac{{12}}{{25}}} \right) - \frac{{12}}{5} - \frac{{12}}{5} = 0$.
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