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If $\cos^{-1}x - \tan^{-1}\frac{\sqrt{1-x^2}}{x}=\pi$, then what are all the possible values of $x$? Please explain in detail.

Irregular User
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Ujjwal
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2 Answers2

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The equation is defined for $x\neq 0$. Set $\theta=\arccos x$. This means $\cos \theta=x,\enspace 0\le x\le \pi,\enspace\theta\neq\dfrac\pi2$.

Now, $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-x^2}\;$ since $\sin\theta\ge 0$ on $[0,\pi]$. Thus $$\tan\theta=\frac{\sqrt{1-x^2}}x,\enspace\text{whence}\enspace\arctan\frac{\sqrt{1-x^2}}x\equiv\theta\mod\pi. $$

As $\;\arctan t\in \Bigl(-\dfrac\pi2,\dfrac\pi2\Bigr)$ by definition, two cases are possible:

  • Either $0\le\theta <\dfrac\pi2$ (corresponding to $x>0$). Then $\;\arctan\dfrac{\sqrt{1-x^2}}x=\theta$ and the equation reduces to $0=\pi$, which has no solution
  • Or $\dfrac\pi2<\theta\le\pi\;$ $\;(x<0$). Then $\;\arctan\dfrac{\sqrt{1-x^2}}x=\theta-\pi$, and the equation reduces to $\pi=\pi$, which is always valid. Hence the solutions are $\;\color{red}{0<x\le \pi}$.
Bernard
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  • But the answer is closed interval [-1,0].please explain further – Ujjwal May 31 '16 at 17:44
  • @Ujjwal: I had inadvertently changed the $-$ sign in the equation with a $+$, sorry! I changed my answer. However, note the value $0$ is not correct. – Bernard May 31 '16 at 18:05
  • It comes from the definition of the $\arctan$ function. I've added an explanation. Is that clearer now? – Bernard May 31 '16 at 18:09
  • In case 1 how did you get 0= $pi$? And In case 2 how why did you take arctan$\frac{\sqrt{1-x^2}}{x}$ as $theta$-$pi$ and not $pi$ - $theta$? And after that how did you get $pi$=$pi$? And does narsimham's explanation convince you? Because according to him 0 should come in the answer. – Ujjwal May 31 '16 at 18:25
  • IN case $1$, I replaced the left-hand side with $\theta_\theta$ (which is $0$, if I'm not mistaken…). In case 2, you cant, because it has to be congruent modulo $\pi$ to $\theta$, not to $-\theta$. As to $\pi=\pi$, I simply replaced again in the left-hand side. – Bernard May 31 '16 at 18:33
  • I'm simply clueless like how can you replace them and what is congruent modulo in case 2 – Ujjwal Jun 01 '16 at 01:17
  • @Ujjwal: In case 2, the l.h.s. becomes $\theta-(\theta-\pi)$, i.e. $\pi$, whence the equality. For your other point, see the third line in my answer: $;\arctan\frac{\sqrt{1-x^2}}x\equiv\color{red}{\theta}\mod\pi$ and by definition $-\theta+\pi\equiv \color{red}-\theta\mod\pi$ – Bernard Jun 01 '16 at 08:51
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As $0\le\cos^{-1}x\le\pi$

As $\tan^{-1}\dfrac{\sqrt{1-x^2}}x=\cos^{-1}x-\pi,$

$-\pi\le\tan^{-1}\dfrac{\sqrt{1-x^2}}x\le0\implies-\dfrac\pi2\le\tan^{-1}\dfrac{\sqrt{1-x^2}}x\le0$ as $\tan^{-1}y\in\Bigl(-\dfrac\pi2,\dfrac\pi2\Bigr)$

$\implies x<0$

Using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,

$\tan^{-1}\dfrac{\sqrt{1-x^2}}x=\cos^{-1}x-\pi=-\cos^{-1}(-x)$

Let $y=-x>0,$

$\implies\cos^{-1}y=-\tan^{-1}\dfrac{\sqrt{1-y^2}}{-y}=\tan^{-1}\dfrac{\sqrt{1-y^2}}y$

Let $\cos^{-1}y=u\implies0\le u<\dfrac\pi2$ and $\cos u=y$

and $\dfrac{\sqrt{1-y^2}}y=\tan u\implies\tan^{-1}\dfrac{\sqrt{1-y^2}}y=u=\cos^{-1}y$

So, the only requirement is $x<0$

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lab bhattacharjee
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