Question : $\triangle PQR$ is right angled at Q. If PQ = 12 cm and PR = 13 cm, find $\tan P+\cot R$.
Option 1: $\frac{9}{10}$
Option 2: $0$
Option 3: $\frac{10}{12}$
Option 4: $\frac{12}{10}$
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Correct Answer: $\frac{10}{12}$
Solution : PQ = 12 cm PR = 13 cm By Pythagoras theorem, PR2 = PQ2 + QR2 132 = 122 + QR2 QR2 = 169 – 144 = 25 ⇒ QR = 5 cm $\tan P$ = $\frac{5}{12}$ ⇒ $\cot R$ = $\frac{5}{12}$ ⇒ $\tan P+\cot R$ = $\frac{5}{12}$+$\frac{5}{12}$ = $\frac{10}{12}$ Hence, the correct answer is $\frac{10}{12}$.
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Question : In a right-angled $\triangle PQR$, PQ = 5 cm, QR = 13 cm and $\angle P=90^{\circ}$. Find the value of $\tan Q-\tan R$.
Option 1: $\frac{5}{14}$
Option 2: $\frac{119}{60}$
Option 3: $\frac{60}{119}$
Option 4: $\frac{14}{5}$
Question : If in a right-angled $\triangle P Q R$, $\tan Q=\frac{5}{12}$, then what is the value of $\cos Q$?
Option 1: $\frac{12}{5}$
Option 2: $\frac{13}{5}$
Option 3: $\frac{12}{13}$
Option 4: $\frac{5}{13}$
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 3: $\frac{18}{5}$
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: $\frac{12}{13}, \frac{5}{13}$
Question : $\triangle PQR$ is right angled at Q. If $m\angle R=60^{\circ}$, what is the length of PR(in cm), If RQ = $4\sqrt3$ cm?
Option 1: $8$
Option 2: $4$
Option 3: $\frac{8}{\sqrt3}$
Option 4: $8\sqrt3$
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