Question
If $\tan\text{A}=\frac{5}{12},$ find the value of $(\sin A + \cos A) \sec A.$
✓
Answer
$\tan\text{A}=\frac{5}{12}=\frac{\text{Perpendicular}}{\text{Base}}$
Draw a right $\triangle\text{ABC}$ in which
$\angle\text{B}=90^\circ\text{AB}=12,\text{BC}=5\text{ units}$
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$=(12)^2+(5)^2=144+25$
$=169=(13)^2$
$\therefore\text{AC}=13\text{ units}$
Now, $\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{5}{13}$
$\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12}{13}$
$\sec\text{A}=\frac{1}{\cos\text{A}}=\frac{13}{12}$
Now $(\sin\text{A}+\cos\text{A})\sec\text{A}$
$=\Big(\frac{5}{13}+\frac{12}{13}\Big)\times\frac{13}{12}$
$=\frac{17}{13}\times\frac{13}{12}=\frac{17}{12}$
Draw a right $\triangle\text{ABC}$ in which
$\angle\text{B}=90^\circ\text{AB}=12,\text{BC}=5\text{ units}$
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$=(12)^2+(5)^2=144+25$
$=169=(13)^2$
$\therefore\text{AC}=13\text{ units}$
Now, $\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{5}{13}$
$\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{12}{13}$
$\sec\text{A}=\frac{1}{\cos\text{A}}=\frac{13}{12}$
Now $(\sin\text{A}+\cos\text{A})\sec\text{A}$
$=\Big(\frac{5}{13}+\frac{12}{13}\Big)\times\frac{13}{12}$
$=\frac{17}{13}\times\frac{13}{12}=\frac{17}{12}$
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