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.