Question

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

Answer 1

We have,

$\cos \theta =\frac{1}{2}$

We know that

$\cos \theta =\frac{Base}{Hypotenuse}$

$Hypotenuse=2$

$Base=1$

We know that

$Height=\sqrt{{\left(Hypotenuse\right)}^2-{\left(Base\right)}^2}$

$Height=\sqrt{4-1}=\sqrt{3}$

So,

$\sec \theta =\frac{1}{\cos \theta }=2$

$\tan \theta =\frac{Height}{Base}=\sqrt{3}$

Since,

$\frac{2\sec \theta }{1+{\tan }^2\theta }$

$=\frac{2\times 2}{1+{\left(\sqrt{3}\right)}^2}$

$=\frac{4}{1+3}$

$=\frac{4}{4}=1$

Hence, this is the answer.