Question

For what interval of variation of $x$, the identity arc $\cos \frac{1-x^2}{1+x^2}=-2$ arc $\tan x$ is true?

• A. $-\infty <x\le 0$
• B. $1<x<\infty$
• C. $0\le x\le 1$
• D. $0\le x<\infty$

Answer 1

Answer: (A)

$-\infty <x\le 0$

Given
$arc\cos \frac{1-x^2}{1+x^2}=-2arc\tan x$
We know that
$arc\cos \frac{1-x^2}{1+x^2}=2arc\tan x$ is true for $0\le x<\infty$
$\therefore$ $arc\cos \frac{1-x^2}{1+x^2}=-2arc\tan x$ is true for $-\infty <x\le 0$