Slight help with inverse trigonometry question

Question

I apologize for the lack of LaTeX, i will try to learn LaTeX and update this question as soon as possible.

I am having some trouble with an inverse trigonometry question and was hoping that someone could give me a hint in the right direction. The question is to simplify $\displaystyle\arctan\Big(\frac{\sqrt{1-x^2}-1}{x}\Big)$

I've tried putting $x$ as $\sec \theta$ and $\tan \theta$ but ran into dead ends. I was hoping someone could provide a hint

Answer 1

Using the defintion of Principal values of inverse Trigonometric functions,

If $2\theta=\arcsin x\implies x=\sin2\theta,-\dfrac\pi2\le2\theta\le\dfrac\pi2$

$\sqrt{1-x^2}=+\cos2\theta$

$F=\dfrac{\sqrt{1-x^2}-1}x=\dfrac{\cos2\theta-1}{\sin2\theta}$

If $\sin\theta=0\iff\theta =0,x=0; F$ is undefined

Otherwise, $F=\dfrac{-2\sin^2\theta}{2\sin\theta\cos\theta}=-\tan\theta=\tan(-\theta)$

So, $\arctan\dfrac{\sqrt{1-x^2}-1}x=\arctan\tan(-\theta)=-\theta=-\dfrac{\arcsin x}2$

Answer 2

If $\arctan\dfrac{\sqrt{1-x^2}-1}x=y, \dfrac{\sqrt{1-x^2}-1}x=\tan y$

$\cos y=\cos y\sqrt{1-x^2}-x\sin y=\cos\left(y+\arccos\sqrt{1-x^2}\right)$

$y=2m\pi\pm\left(y+\arccos\sqrt{1-x^2}\right)$

$+\implies y=2m\pi+\left(y+\arccos\sqrt{1-x^2}\right)\iff\arccos\sqrt{1-x^2}=2m\pi\implies\sqrt{1-x^2}=\cos(-2m\pi)=1\implies x=0$ which makes $\dfrac{\sqrt{1-x^2}-1}x$ undefined

$-\implies y=2m\pi-\left(y+\arccos\sqrt{1-x^2}\right)$ $\iff y=m\pi-\dfrac{\arccos\sqrt{1-x^2}}2=m\pi-\dfrac{\arcsin x}2$

As $-\dfrac\pi2\le\arcsin x\le\dfrac\pi2\iff -\dfrac\pi4\le\dfrac{\arcsin x}2\le\dfrac\pi4\iff -\dfrac\pi4\le-\dfrac{\arcsin x}2\le\dfrac\pi4$

$\iff m\pi-\dfrac\pi4\le m\pi-\dfrac{\arcsin x}2\le m\pi+\dfrac\pi4$

As $-\dfrac\pi2\le y\le\dfrac\pi2, m=0$