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

Answer 1

Answer: (B)

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