Question

If $f\left(x\right)=\frac{x}{\sqrt{1-x^2}},g\left(x\right)=\frac{x}{\sqrt{1+x^2}}$ then $(f\circ g)\left(x\right)=$

• A. $\frac{x}{\sqrt{1-x^2}}$
• B. $\frac{x}{\sqrt{1+x^2}}$
• C. $\frac{1-x^2}{\sqrt{1-x^2}}$
• D. $x$

Answer 1

The correct option is D $x$

Let $x=\tan \theta$
$(f\circ g)\left(x\right)=f\left[g\right(x\left)\right]=f\left[g\right(\tan \theta \left)\right]$.
Since $g\left(x\right)=\frac{x}{\sqrt{1+x^2}}$, we have
$f\left[g\right(\tan \theta \left)\right]=f\left[\frac{\tan \theta }{\sqrt{1+{\tan }^2\theta }}\right]=f\left[\frac{\tan \theta }{\sec \theta }\right]=f\left(\sin \theta \right)$
Also since $f\left(x\right)=\frac{x}{\sqrt{1-x^2}}$ , we have
$f\left(\sin \theta \right)=\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$
$=\tan \theta =x$