17 Views

Question : If $4\left(\operatorname{cosec}^2 57^{\circ}-\tan ^2 33^{\circ}\right)-\cos 90^{\circ}-y \tan ^2 66^{\circ} \tan ^2 24^{\circ}=\frac{y}{2}$, the value of $y$ is:

Option 1: $\frac{8}{3}$

Option 2: $\frac{3}{8}$

Option 3: $8$

Option 4: $\frac{1}{3}$


Team Careers360 10th Jan, 2024
Answer (1)
Team Careers360 20th Jan, 2024

Correct Answer: $\frac{8}{3}$


Solution : $\operatorname{cosec}(90^{\circ} - \theta) = \sec \theta$
$\tan\theta = \frac{1}{\cot\theta}$
So, $4\left(\operatorname{cosec}^2 57^{\circ}-\tan ^2 33^{\circ}\right)-\cos 90^{\circ}-y \tan ^2 66^{\circ} \tan ^2 24^{\circ}=\frac{y}{2}$
⇒ $(\operatorname{cosec}^{2}(90-33)^{\circ} - \tan^{2}33^{\circ}) - 0- y×\tan^{2}66^{\circ} × \tan^{2}(90-66)^{\circ}$ = $\frac{y}{2}$ 
⇒ $4(\sec^{2}57^{\circ}- \tan^{2}33^{\circ}) -  y × \tan^{2}66^{\circ} × \cot^{2}66^{\circ} = \frac{y}{2}$
⇒ $4 - y = \frac{y}{2}$
⇒ y = $\frac{8}{3}$
Hence, the correct answer is $\frac{8}{3}$.

Know More About

Related Questions

Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
View All Application Forms

Upcoming Exams

Tier II Admit Card Date: 7 Apr, 2026 - 10 Apr, 2026
Preliminary Exam Admit Card Date: 2 Apr, 2026 - 13 Apr, 2026
Preliminary Exam Exam Date: 24 May, 2026 - 24 May, 2026

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books