Question

If $A$ lies between ${180}^0$ and ${270}^0$ and $12\tan A=5$, then find the value of $5\cot A-13\sin A+12\sec A$.

Answer 1

$12\tan A=5$ i.e. $\tan =A\frac{5}{12}$
$\therefore \cot A=\frac{12}{5},\sin A=\frac{5}{13}$ and $\sec A=\frac{13}{12}$
Thus $5\cot A-13\sin A+12\sec A$
$=5\times \frac{12}{5}+13\times \frac{5}{13}-12\times \frac{13}{12}$
$=12+5-13=4$