solve $\tan(x) = \sqrt{1-x^2}$

Question

I am not sure if you should be deriving it or converting tan into $\sin(x)/\cos(x)$. Even then, I do not know what to do from there

Answer 1

For $|x| > 1$, $\sqrt{1-x^2}$ is undefined, so there are no solutions in this range.

For $-1 \le x < 0$, we have that $\tan x < 0 < \sqrt{1-x^2}$, so there are no solutions in this range.

For $0 \le x \le 1$, we have that $\tan x$ is increasing and $\sqrt{1-x^2}$ is decreasing, so there is at most one solution. Also, by the intermediate value theorem, we see there must be at least one solution. Therefore, there is exactly one solution $x \in [0,1]$.

Using WolframAlpha, it appears the solution is $x \approx 0.649888946665696\ldots$. I don't think there is a nice closed form for this.

Answer 2

The tangent function has an s-shaped curve running between two vertical asymptotes at $-\frac{\pi}2$ and $+\frac{\pi}2$ and through the origin (with slope $1$). The right-hand side expression $\sqrt{1-x^2}$ is the upper half circle around the origin, with radius $1$.

There is a single intersection between these curves, located in the vicinity of $(\frac1{\sqrt2}, \frac1{\sqrt2})$ (obtained from the crude approximation $\tan x\approx x$).

A better approximation can be obtained by rewriting $tan^2x+x^2-1=0$, and using the Taylor development $tan^2x\approx x^2+\frac23x^4$. This gives the solution $x\approx\sqrt{\frac{\sqrt{15}-3}2}\approx0.661$.

Newton will quickly polish this root.

0.660675164588
0.650054962499
0.649888986250
0.649888946666
0.649888946666
0.649888946666