Question

the formula $\cos {}^{-1}\frac{1-x^2}{1+x^2}=2{\tan }^{-1}xholdsfor:$

• A. $x\in R$
• B. $|x|⩽1$
• C. $x\in (-1,1]$
• D. $x\in [1,+\infty )$

Answer 1

The correct option is A $x\in R$

Solution :- We know by general solution that

$ta{n}^{-1}x=\theta$ such that $\infty <x<\infty$

$\therefore x=tan$ $\theta x\in R$

Now using double angle property

$cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$

$\Rightarrow cos2\theta =\frac{1-x^2}{1+x^2}$ where $x\in R$

$\Rightarrow 2\theta =co{s}^{-1}\frac{1-x^2}{1+x^2}$ where $x\in R$

$\Rightarrow 2ta{n}^{-1}x=co{s}^{-1}\frac{1-x^2}{1+x^2}$ where $x\in R$

$\therefore$ For formula $2ta{n}^{-1}x=co{s}^{-1}\frac{1-x^2}{1+x^2}$

x holds for $x\in R$ Ans