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}} $
$\Rightarrow \sec \left(\tan ^{-1} \frac{y}{2}\right)=\frac{\sqrt{4+y^{2}}}{2}$