Find inverse of a function $t = \frac{1}{\sqrt{1+x^2}}$

Question

I have a formula $t = \frac{1}{\sqrt{1+x^2}}$

How is it possible to convert it into $x = +-\frac{\sqrt{1-t^2}}{t}$

I am assuming that it is an inverse function that is calculated by replacing x with t in the original equation, and then solving for x?

But I can't figure out how did it become like this. Am I just failing with the basic algebra, or is there something else to it?

Answer 1

Square both sides to get $$ t^2=\frac {1}{1+x^2}$$ Reciprocate to get $$ \frac {1}{t^2}=1+x^2$$ $$x^2=\frac {1}{t^2}-1=\frac{1-t^2}{t^2}$$

Take square root and switch $x$ and $t$

Answer 2

We have $0<t\le 1$ and then

$$t = \frac{1}{\sqrt{1+x^2}} \iff 1+x^2=\frac1{t^2} \iff x^2=\frac{1-t^2}{t^2}$$

and then assuming $x\ge 0$

$$x=\frac{\sqrt{1-t^2}}{t}$$

or as an alternative, assuming $x\le 0$

$$x=-\frac{\sqrt{1-t^2}}{t}$$

Answer 3

$t^2 =\frac{1}{1+x^2}$ $\implies 1+x^2 =\frac{1}{t^2}$

                   $ \implies x^2=\frac{1}{t^2 }-1$

i.e $x^2=\frac{1-t^2}{t^2} \implies $ $x = +-\frac{\sqrt{1-t^2}}{t}$

Answer 4

$t = \frac{1}{\sqrt{1+x^{2}}}$

$t^{2} = \frac{1}{1+x^{2}}$

$1+x^{2} = \frac{1}{t^{2}}$

$x^{2} = \frac{1-t^{2}}{t^{2}}$

$x = \pm \sqrt{\frac{1-t^{2}}{t^{2}}} = \pm \frac{\sqrt{1-t^{2}}}{t}$ , since t > 0

Answer 5

$t = \frac{1}{\sqrt{1+x^2}}$

$t\sqrt{1+x^2} = 1$

$\sqrt{1+x^2} = \frac{1}{t}$

$1 + x^2 = \frac{1}{t^2}$

$x^2 = \frac{1}{t^2} - 1$

$x^2 = \frac{1 - t^2}{t^2)}$

$x = \pm\sqrt{\frac{1-t^2}{t^2}}$

$x = \pm\frac{\sqrt{1-t^2}}{t}$

Answer 6

Let us transform the initial equation and try to preserve the original solutions: \begin{align} t &= \frac{1}{\sqrt{1 + x^2}} \iff \\ t^2 &= \frac{1}{1+x^2} \quad \wedge \quad t > 0 \iff \\ 1 + x^2 &= \frac{1}{t^2} \quad \wedge \quad t > 0 \iff \\ x^2 &= \frac{1}{t^2} - 1 \quad \wedge \quad t > 0 \iff \\ x &= \pm\sqrt{\frac{1}{t^2} - 1} \quad \wedge \quad t > 0 \iff \\ \end{align}

Now let us check it:

(Large version)

The green curve is the original function, with $t$ on the $y$ axis.

The blue curve is just the relation of the last line, positive root, both signs for $t$. The orange curve is the last line, negative root, both signs for $t$.

$t$ is here on the $x$-axis, and we see, mirroring at the line $y=x$, that indeed $t > 0$ is the correct choice for $t$.