A) \[-1\]
B) \[5\]
C) \[\frac{1}{2}\]
D) \[1\]
E) \[-\frac{1}{2}\]
Correct Answer: D
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.
You need to login to perform this action.
You will be redirected in
3 sec
