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