Question : $\triangle PQR$ is a right-angled triangle. $\angle Q = 90^\circ$, PQ = 12 cm, and QR = 5 cm. What is the value of $\operatorname{cosec}P+\sec R$?
Option 1: $\frac{26}{5}$
Option 2: $\frac{13}{5}$
Option 3: $\frac{18}{5}$
Option 4: $\frac{14}{5}$
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Correct Answer: $\frac{26}{5}$
Solution :
As $PR^2=PQ^2+QR^2$
⇒ $PR^2=12^2+5^2$
⇒ $PR^2=144+25$
⇒ $PR^2=169$
⇒ $PR=\sqrt{169}$
⇒ $PR=13$ cm
Now, $\operatorname{cosec} \ P=\frac{PR}{QR}=\frac{13}{5}$
and $\sec R=\frac{PR}{QR}=\frac{13}{5}$
$\therefore \operatorname{cosec} \ P+\sec R=\frac{13}{5}+\frac{13}{5}=\frac{26}{5}$
Hence, the correct answer is $\frac{26}{5}$.
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