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 $