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)

$1$

Answer 2

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