Question

If $\sin A=\frac{1}{2}$, then find the value of $\frac{2\sec A}{1+{\tan }^{2}A}$.

Answer 1

Consider the given data.

$\sin A=\frac{1}{2}=\frac{Perpendicular}{Hypotenuse}$

Perpendicular $=1unit$, Hypotenuse $=2unit$

Using Pythagoras theorem, we get

Base $=\sqrt{3}unit$

Now,

$\sec A=\frac{Hypotenuse}{Base}=\frac{2}{\sqrt{3}}$

$\tan A=\frac{Perpendicular}{Base}=\frac{1}{\sqrt{3}}$

So,

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

$=\frac{2\times \frac{2}{\sqrt{3}}}{1+\frac{1}{{\sqrt{3}}^2}}$

$=\frac{\frac{4}{\sqrt{3}}}{\frac{4}{3}}$

$=\frac{3}{\sqrt{3}}$

$=\sqrt{3}$

Hence, the value is $\sqrt{3}$.