Question

If $f\left(y\right)=\frac{y}{\sqrt{1-y^2}}$; $g\left(y\right)=\frac{y}{\sqrt{1+y^2}}$ then $\left(fog\right)y$ is equal to

• A. $\frac{y}{\sqrt{1-y^2}}$
• B. $\frac{y}{\sqrt{1+y^2}}$
• C. $y$
• D. $2f\left(x\right)$

Answer 1

Answer: (C)

$y$

Given $f\left(y\right)=\frac{y}{\sqrt{1-y^2}}$ and $g\left(y\right)=\frac{y}{\sqrt{1+y^2}}$
$\therefore \left(fog\right)y=f\left(g\right(y\left)\right)=\frac{\frac{y}{\sqrt{1+y^2}}}{\sqrt{1-\frac{y^2}{1+y^2}}}=y$