Question

If ${\tan }^{-1}(1-x),{\tan }^{-1}(x)$ and ${\tan }^{-1}(1+x)$ are in $A.P.$, then the value of $x^3+x^2$ is equal to

• A. $2$
• B. $-1$
• C. $1$
• D. $-2$

Answer 1

Answer: (C)

Given that, ${\tan }^{-1}(1-x),{\tan }^{-1}x$
and ${\tan }^{-1}(1+x)$ are in AP, then
$2{\tan }^{-1}x={\tan }^{-1}(1-x)+{\tan }^{-1}(1+x)$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)={\tan }^{-1}\left(\frac{1-x+1+x}{1-(1-x)(1+x)}\right)$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)={\tan }^{-1}\left(\frac{2}{x^2}\right)$
$\Rightarrow$ $\frac{2x}{1-x^2}=\frac{2}{x^2}$
$\Rightarrow$ $x^3=1-x^2$
$\Rightarrow$ $x^3+x^2=1$