Question

$\sec \left({\tan }^{-1}\frac{y}{2}\right)=$

• A. $\sqrt{\frac{4+y^2}{2}}$
• B. $\sqrt{\frac{4-y^2}{2}}$
• C. $\frac{\sqrt{4+y^2}}{2}$
• D. $\frac{\sqrt{4-y^2}}{2}$

Answer 1

Answer: (C)

We have,
$\sec \left\{{\tan }^{-1}\left(\frac{y}{2}\right)\right\}$
Let${\tan }^{-1}\frac{y}{2}=x$
$\therefore \frac{y}{2}=\tan x$
Squaring on both sides, we get
$\Rightarrow \frac{y^2}{4}={\tan }^{2}x$
$\Rightarrow \frac{y^2}{4}={\sec }^{2}x-1$
$\Rightarrow {\sec }^{2}x=1+\frac{y^2}{4}$
$\Rightarrow \sec x=\sqrt{\frac{4+y^2}{4}}$
$\Rightarrow \sec \left({\tan }^{-1}\frac{y}{2}\right)=\frac{\sqrt{4+y^2}}{2}$