Question

If $x\in (0,1)$, then the expression $\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)$ simplifies to:

• A. ${\cot }^{-1}\sqrt{x}$
• B. ${\tan }^{-1}\sqrt{x}$
• C. ${\cos }^{-1}\sqrt{x}$
• D. $-{\tan }^{-1}\sqrt{x}$

Answer 1

Answer: (B), (D)

$-{\tan }^{-1}\sqrt{x}$

Let $x={\tan }^2\theta$
For $x\in (0,1),$ we have
$\theta \in (-\frac{\pi }{4},0)\cup (0,\frac{\pi }{4})$
Now, the expression can be written as
$\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)=\frac{1}{2}{\cos }^{-1}\left(\cos 2\theta \right)$
As $2\theta \in (-\frac{\pi }{2},\frac{\pi }{2})-\left\{0\right\}$
In the given symmetrical interval, we get
${\cos }^{-1}\left(\cos 2\theta \right)=\{\begin{array}{c}2\theta ,2\theta \in (0,\frac{\pi }{2})\\ -2\theta ,2\theta \in (-\frac{\pi }{2},0)\end{array}$
Therefore
$\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)=\pm \theta =\pm {\tan }^{-1}\sqrt{x}$