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  4. If tan -1(x+2)+ tan -1(x-2)- tan -1( (1/2) )=0, then one of the values of x is equal to

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Q. If $ {{\tan }^{-1}}(x+2)+{{\tan }^{-1}}(x-2)-{{\tan }^{-1}}\left( \frac{1}{2} \right)=0, $ then one of the values of $ x $ is equal to

KEAMKEAM 2010Inverse Trigonometric Functions

Solution:

Given that, $ {{\tan }^{-1}}(x+2)+{{\tan }^{-1}}(x+2)-{{\tan }^{-1}}\left( \frac{1}{2} \right)=0 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{x+2+x-2}{1-({{x}^{2}}-4)} \right)-{{\tan }^{-1}}\left( \frac{1}{2} \right)=0 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{2x}{1-({{x}^{2}}-4)} \right)-{{\tan }^{-1}}\left( \frac{1}{2} \right)=0 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{\left( \frac{2x}{1-({{x}^{2}}-4)}-\frac{1}{2} \right)}{1+\frac{2x}{(1-({{x}^{2}}-4))}.\frac{1}{2}} \right)=0 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{\frac{4x-1+{{x}^{2}}-4}{2(1-({{x}^{2}}-4))}}{\frac{2(1-({{x}^{2}}-4))+2x}{2(1-({{x}^{2}}-4))}} \right)=0 $
$ \Rightarrow $ $ {{\tan }^{-1}}\left( \frac{{{x}^{2}}+4x-5}{-2{{x}^{2}}+2x+6} \right)=0 $
$ \Rightarrow $ $ {{x}^{2}}+4x-5=0 $
$ \Rightarrow $ $ x=1,\,-5 $
$ \Rightarrow $ $ x=1 $ $ \because $ $ -5 $ is not possible.