Question : Let ABC be a triangle right-angled at B. If $\tan A = \frac{12}{5}$, then find the values of $\operatorname{cosec A}$ and $\sec A$, respectively.
Option 1: $\frac{13}{10}, \frac{5}{13}$
Option 2: $\frac{13}{12},\frac{13}{5}$
Option 3: $\frac{10}{13}, \frac{5}{13}$
Option 4: $\frac{12}{13}, \frac{5}{13}$
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Correct Answer: $\frac{13}{12},\frac{13}{5}$
Solution :
Given, $\tan A=\frac{12}{5}$
$\tan A = \frac{\text{perpendicular}}{\text{base}}$
Using pythagoras theorem,
$\small\text{Hypotenuse}^2=\text{Base}^2+\text{Perpendicular}^2$
⇒ $h^2=5^2+12^2$
⇒ $h^2=25+144$
⇒ $h^2=169$
⇒ $h=13$
$\therefore$ $\operatorname{cosec A}=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{13}{12}$
And, $\sec A=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{13}{5}$
Hence, the correct answer is $\frac{13}{12},\frac{13}{5}$.
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