A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) 0
D) \[\frac{9}{4}\]
Correct Answer: A
Solution :
\[\sin [{{\cot }^{-1}}(x+1)]=\sin \left( {{\sin }^{-1}}\frac{1}{\sqrt{{{x}^{2}}+2x+2}} \right)\] \[=\frac{1}{\sqrt{{{x}^{2}}+2x+2}}\] \[\cos ({{\tan }^{-1}}x)=\cos \left( {{\cos }^{-1}}\frac{1}{\sqrt{1+{{x}^{2}}}} \right)=\frac{1}{\sqrt{1+{{x}^{2}}}}\] Thus, \[\frac{1}{\sqrt{{{x}^{2}}+2x+2}}=\frac{1}{\sqrt{1+{{x}^{2}}}}\] \[\Rightarrow {{x}^{2}}+2x+2=1+{{x}^{2}}\]\[\Rightarrow \] \[x=-\frac{1}{2}\].
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