Question

If $\cos \theta =\frac{1}{2}$, find the value of $\frac{2\sec \theta }{1+{\tan }^2\theta }$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

Given, $\cos \theta =\frac{1}{2}$
Therefore, ${\sin }^2\theta =1-{\cos }^2\theta$
$=1-{\left(\frac{1}{2}\right)}^2=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \sin \theta =\pm \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{2}$, taking only the positive values
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{2}{1}=\sqrt{3}$
and $\sec \theta =2$
Therefore, the value of $\frac{2\sec \theta }{1+{\tan }^2\theta }=\frac{2\times 2}{1+3}=\frac{4}{4}=1$.