Question
Given : $5 \cos A - 12 \sin A = 0;$ evaluate:$\frac{\sin A+\cos A}{2 \cos A-\sin A}$
✓
Answer
$5 \cos A – 12 \sin A = 0$
$5 \cos A = 12 \sin A$
$ \frac{\sin A}{\cos A}=\frac{5}{12}$
$\tan A=\frac{5}{12} $
Now,
$\frac{\sin A+\cos A}{2 \cos A-\sin A}=\frac{\frac{\sin A}{\cos A}+\frac{\cos A}{\cos A}}{2 \frac{\cos A}{\cos A}-\frac{\sin A}{\cos A}}$
$=\frac{\tan A +1}{2-\tan A }$
$=\frac{\frac{5}{12}+1}{2-\frac{5}{12}}$
$=\frac{\frac{17}{12}}{\frac{19}{12}}$
$=\frac{17}{19}$
$5 \cos A = 12 \sin A$
$ \frac{\sin A}{\cos A}=\frac{5}{12}$
$\tan A=\frac{5}{12} $
Now,
$\frac{\sin A+\cos A}{2 \cos A-\sin A}=\frac{\frac{\sin A}{\cos A}+\frac{\cos A}{\cos A}}{2 \frac{\cos A}{\cos A}-\frac{\sin A}{\cos A}}$
$=\frac{\tan A +1}{2-\tan A }$
$=\frac{\frac{5}{12}+1}{2-\frac{5}{12}}$
$=\frac{\frac{17}{12}}{\frac{19}{12}}$
$=\frac{17}{19}$
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