8 Views

Question : If $\sin \beta=\frac{1}{3},(\sec \beta-\tan \beta)^2$ is equal to:

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

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

Option 3: $\frac{1}{2}$

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


Team Careers360 5th Jan, 2024
Answer (1)
Team Careers360 20th Jan, 2024

Correct Answer: $\frac{1}{2}$


Solution : Given:
$\sin \beta=\frac{1}{3}$
We know that
$ \cos \beta\ =\ \sqrt{1\ -\ \sin^{2}\beta}\ =\ \sqrt{1\ -\ \frac{1}{9}}\ =\ \frac{\sqrt{8}}{3}$
$\Rightarrow \sec \beta\ =\ \frac{1}{\cos \beta}\ =\ \frac{3}{\sqrt{8}}$
$\Rightarrow \tan \beta\ =\ \frac{\sin\beta}{\cos \beta}\ =\ \frac{\frac{1}{3}}{\frac{\sqrt{8}}{3}}\ =\ \frac{1}{\sqrt{8}}$
Now,
$(\sec \beta-\tan \beta)^2$
$=\ \left ( \frac{3}{\sqrt{8}}\ -\ \frac{1}{\sqrt{8}} \right )^2$
$=\ \left ( \frac{2}{\sqrt{8}} \right )^2$
$=\ \frac{1}{2}$
Hence, the correct answer is $\frac{1}{2}$.

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