Question

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

Answer 1

Let us consider the problem:

Given: $\sin A=\frac{1}{2}=\frac{perpendicular}{hypotenus}$

to calculate the value $\frac{2\sec A}{1+{\tan }^{2}A}$

by using Pythagoras theorem:

base$=$$\sqrt{{\left(hypotenus\right)}^2-{\left(perpenclicular\right)}^2}$

therefore, $cosA$

$=\frac{base}{hypotnus}=\frac{\sqrt{3}}{2}$

let us consider

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

$\Rightarrow$ $\frac{2\sec A}{{\sec }^{2}A}\left(\mathrm{since}1+ta{n}^{2}a={\sec }^{2}a\right)$

$=\frac{2}{\sec A}$

$=2\cos A\left(\mathrm{since}\frac{1}{\sec A}=\cos a\right)$

$=2\times \frac{\sqrt{3}}{2}$

$=\sqrt{3}$