Inverse Trigonometric Functions Set-2

$\dfrac{3}{5}$

$\dfrac{24}{25}$

$\dfrac{7}{25}$

$\dfrac{16}{25}$

$\displaystyle \frac{-3-\sqrt{17}}{4}$

$\displaystyle \frac{-3-\sqrt{17}}{8}$

$\displaystyle \frac{2}{3}$

$\displaystyle \frac{-3+\sqrt{17}}{4}$

$\dfrac{3}{2}$

$1$

$\dfrac{6}{5}$

$\dfrac{5}{3}$

$x\in\mathbb{R}\setminus\{\pm1\}$

$x\in(-\infty,-1)\cup(1,\infty)$

$x\in(-1,1)$

$\{0\}$

$0$

$\dfrac{1}{2}$

$-\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{7}{25}$

$\dfrac{12}{25}$

$\dfrac{24}{25}$

$\dfrac{9}{25}$

$x=\dfrac{1}{2}$

$x=-\dfrac{1}{2}$

$x=0$

$x=-\dfrac{1}{4}$

$x=\dfrac{-3-\sqrt{17}}{4}$

$x=\dfrac{-3+\sqrt{17}}{4}$

Both of the above

No real solution

$2\arctan x\quad\text{for }|x|\le1,\quad \pi-2\arctan x\quad\text{for }x>1,\quad -\pi-2\arctan x\quad\text{for }x<-1$

$2\arctan x\quad\text{for }|x|\le1,\quad 2\pi-2\arctan x\quad\text{for }x>1,\quad -2\pi-2\arctan x\quad\text{for }x<-1$

$2\arctan x\quad\text{for all real }x$

$2\arctan x\quad\text{for }|x|\le1,\quad \pi-2\arctan x\quad\text{for }x>1,\quad -\pi+2\arctan x\quad\text{for }x<-1$

$2\arctan|x|$

$2\arctan x$

$\begin{aligned} &2\arctan x\quad\text{for }x\ge0,\\ &\pi-2\arctan x\quad\text{for }x<0 \end{aligned}$

$\pi-2\arctan|x|$