Question

If $A$ lies between ${180}^0$ and ${270}^0$ and $3\tan A=4$, then find the value of 2$\cot A-\cos A+\sin A$.

Answer 1

$3\tan A=4\tan A=\frac{4}{3}$
$\therefore \sin A=\frac{4}{5}and\cos A=\frac{3}{5}$
$\therefore$ A lies in the third quadrant so
$\sin A=-\frac{4}{5}and\cos A=-\frac{3}{5}$
Thus
$2\cot A-5\cos A+\sin A$
$=2\times \frac{3}{4}-5\times (-\frac{3}{5})-\frac{4}{5}$
$=\frac{3}{2}+3-\frac{4}{5}=3\frac{7}{10}$