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: $\frac{14}{5}$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $\frac{119}{60}$
Solution :
Given: PQ = 5 cm
QR = 13 cm
$\angle P=90^{\circ}$
In $\triangle PQR$,
$RQ^2=PQ^2+PR^2$
⇒ $13^2=5^2+PR^2$
⇒ $PR^2=169-25$
⇒ $PR^2=144$
$\therefore PR = 12$
Now, $\tan Q-\tan R$
$=\frac{12}{5}-\frac{5}{12}$
$=\frac{144-25}{60}$
$=\frac{119}{60}$
Hence, the correct answer is $\frac{119}{60}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates BrochureYour Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.