Question : If $\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}=\frac{4}{5}$, then the value of $\frac{\operatorname{cosec}^2 \theta}{2-\operatorname{cosec}^2 \theta}$ is:
Option 1: $\frac{16}{25}$
Option 2: $\frac{40}{41}$
Option 3: $\frac{41}{40}$
Option 4: $\frac{31}{30}$
New: SSC CHSL Tier 2 answer key released | 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: $\frac{41}{40}$
Solution :
Given: $\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}=\frac{4}{5}$
⇒ $5\sin \theta - 5\cos \theta = 4\sin \theta + 4\cos \theta$
$\sin \theta = 9\cos \theta$
Dividing both sides by $\cos \theta$, we get:
$\tan \theta = 9$
Now, we know that:
$\tan\theta=\frac{p}{b}=\frac{9}{1}$
So, $h=\sqrt{p^2+b^2}=\sqrt{9^2+1^2}=\sqrt{82}$
Now, $\frac{\operatorname{cosec}^2 \theta}{2-\operatorname{cosec}^2 \theta}$
⇒ $\frac{(\frac{h}{p})^{2}}{2-(\frac{h}{p})^{2}}$
Putting the values of $p$, $b$ and $h$, we get:
= $\frac{\frac{82}{81}}{2-\frac{82}{81}}$
= $\frac{\frac{82}{81}}{\frac{162-82}{81}}$
= $\frac{82}{80}$
= $\frac{41}{40}$
Hence, the correct answer is $\frac{41}{40}$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Trending Articles around SSC CHSL
