Take the tour ×  Real solutions to   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.enter image description here  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.  share|improve
this question Is there a typo in your solution? As written,
$1-\varphi$ is negative, so it shouldn't be under a radical.
Maybe you me