Compute: $\arcsin x+ \arctan \frac{\sqrt{1-x^2}}{x}$

Question

$$\arcsin x+ \arctan \frac{\sqrt{1-x^2}}{x}$$ This problem is from my final test for this semester in the third year highschool. I think this should be solved by using Lagrange or Rolle. I computed the derivative for $f(x)$ ,$f(x)$=$\arcsin x+ \arctan \frac{\sqrt{1-x^2}}{x}$ $f'(x)=0; $ Using Rolle we know that $f(a)=f(b) $ where $x\in(a,b)$

Answer 1

This isn't a calculus problem at all; it's trigonometry. If you draw a right triangle, with one leg equal to $x$, and with the hypotenuse equal to $1$, then the other leg will have length $\sqrt{1-x^2}$. Your first term represents the angle opposite the $x$ side, the your second term represents the angle adjacent to it. Those two angles add up to a right angle, i.e., $\pi\over 2$.

Of course, this analysis assumes that $x$ is the length of a triangle side, which is to say, that $x>0$. To take care of the other possibility, consider that $\arcsin$ and $\arctan$ are both odd functions.

Answer 2

HINT

Recall that by right triangle properties

$$\tan(\arcsin x)=\frac x {\sqrt{1-x^2}}$$

Answer 3

Let $\arcsin x=y\implies x=\sin y,-\dfrac\pi2\le y\le\dfrac\pi2$

$\arctan\dfrac{\sqrt{1-x^2}}x=\arctan(\cot y)=\arctan\left(\left(\dfrac\pi2-y\right)\right)$

$=\begin{cases}\dfrac\pi2-y&\mbox{if } -\dfrac\pi2\le\dfrac\pi2-y\le\dfrac\pi2\iff\pi\ge y\ge0 \\\dfrac\pi2-y-\pi & \mbox{if } y\le0 \end{cases}$