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