Other ways of solving $\cot^{-1}(x)=\sin^{-1}(x)$ Real solutions to

$$\cot^{-1}(x)=\sin^{-1}(x)$$

I found this problem in an exam years ago and I solved it using geometry. The first mistake I made was assuming $\cot^{-1}(x)=\cfrac{1}{\tan^{-1}(x)}$, which i realized quickly enough. I took a $\theta$ such that $\theta=\sin^{-1}(x)=\cot^{-1}(x)\implies \cot \theta=\sin \theta=x$ . This right angled triangle satisfies the condition.
Using Pythagoras' Theorem on the triangle and one has the equation $$x^4+x^2-1=0\implies x=\pm\sqrt{\varphi-1} \ \quad \text{where} \ \varphi \text { is the Golden Ratio, } \ \varphi = \cfrac{1+\sqrt 5}2 $$
I gave this question to my teacher in school after the exam and he solved it using calculus and stuffs. I can't really remember (It was 6 years ago) but he clearly didn't use my method.
I would like if someone can show me (an)other way(s) of solving this question.
 A: You could apply $\sin$ to both sides:
$$\sin(\cot^{-1}x)=x$$
Now it helps to know how to simplify the left side. If the cotangent of an angle is $x$, and that angle is part of a right triangle, then we can scale the triangle so that $x$ is the adjacent length, $1$ is opposite, and thus $\sqrt{1+x^2}$ is the hypotenuse. Taking $\sin$:
$$\frac{1}{\sqrt{1+x^2}}=x$$
Squaring and clearing denominators is another way to reduce to your equation of $x^4+x^2=1$
A: One could prove it with trigonometric identities (although this is in the same spirit as your geometric route).
Take sines of both sides, to get $\sin(\cot^{-1}(x))=x$. Square both sides and take reciprocals, so:
$$\frac{1}{x^{2}}=\frac{1}{\sin^{2}(\cot^{-1}(x))}=\csc^{2}(\cot^{-1}(x))=1+\cot^{2}(\cot^{-1}(x))=1+x^{2}$$
In other words, $$x^{4}+x^{2}-1=0 \implies x^{2}=\varphi-1$$
A: Given a real $x$ the expression $\cot^{-1}(x)={\rm arccot}(x)$ denotes the unique angle $\alpha\in\ ]0,\pi[\ $ with $\cot\alpha=x$. Similarly, given a real $x\in[{-1},1]$ the expression $\sin^{-1}(x)=\arcsin (x)$ denotes the unique angle $\alpha\in[-{\pi\over2},{\pi\over2}]$ with $\sin\alpha=x$.
Therefore the equation $\cot^{-1}(x)=\sin^{-1}(x)$ talks about a number $x\in[-1,1]$ and implicitly about an  angle $\alpha\in\ ]0,{\pi\over2}]$ such that
$$x=\cot\alpha=\sin\alpha\ .$$
It follows that $\cos\alpha=\sin^2\alpha=1-\cos^2\alpha$, or
$$\cos\alpha={\sqrt{5}-1\over2}\ .$$
From $\sin^2\alpha={\sqrt{5}-1\over2}$ we conclude $x=\sin\alpha=\pm\sqrt{{\sqrt{5}-1\over2}}$; but as $\alpha$ has to lie in the interval $\ ]0,{\pi\over2}]$ the only possible solution is 
$$x=\sqrt{{\sqrt{5}-1\over2}}\doteq 0.78615\ .$$
Since we arrived at this  solution by means of correct but  "one-way" algebraic manipulations, and not in possession of a general theory about equations of the given type, we should test the found $x$ whether it actually solves the original equation. I leave this verification to the OP.
