# $\pm$ sign in $y=\arcsin\frac{x}{\sqrt{1+x^2}}$

If: $$y=\arcsin\frac{x}{\sqrt{1+x^2}}$$ Then: $$\sin(y)=\frac{x}{\sqrt{1+x^2}}$$ $$\cos^2(y)=1-\sin^2(y)=\frac{1}{1+x^2}$$ $$\tan^2(y)=\sec^2(y)-1=1+x^2-1=x^2$$ Therefore I would say: $$\tan(y)=\pm x$$ However, my calculus book says (without the $\pm$): $$\tan(y)=x$$

Question: Why can we remove the $\pm$?

• I think we can write as $tan(y)=|x|$ – Archis Welankar Mar 12 '16 at 15:00
• Essentially because $a=\arcsin(b)$ and $\sin(a)=b$ aren't equivalent statements. The $\arcsin$ function is usually defined in such a way that $\cos(\arcsin(b))\ge0$. – Yves Daoust Mar 14 '16 at 9:00

The sign of $\tan(\theta)$ is not uniquely determined from an equation $\sin(\theta) = a$, which has two solutions with opposite signs for the tangent. Under any convention for choosing one of the two $\theta$'s as the value of $\arcsin(a)$, the tangent is uniquely determined. The convention consistent with what you wrote is $\arcsin \in (-\pi/2, \pi/2]$
• Your answer makes sense, but let's we were to have the convention to chose the value for tangent on $(\pi/2, 3\pi/2]$. If we plug in $\tan y = -x$, we would still get $y = \arcsin (\sin (-y)$. Would this reduce then to $y=y$? – Ovi Mar 12 '16 at 17:02
We can get rid of the $\pm$ sign because in $y=\arcsin\frac{x}{\sqrt{1+x^2}}$, $x$ and $y$ have to have the same sign: For $-\pi/2 < y\le\pi/2$ if $x$ is positive, then $y$ is positive then also $\tan(y)$ is positive. If $x$ is negative then $y$ is negative then $\tan(y)$ is negative.
Therefore, if $\tan^2(y)=x^2$ then $\tan(y)=x$