A) \[1/2\]
B) \[1\]
C) \[0\]
D) \[-1/2\]
Correct Answer: D
Solution :
\[\sin ({{\cot }^{-1}}(x+1))=\cos ({{\tan }^{-1}}x)\] \[\Rightarrow \]\[\cos (\pi /2-{{\cot }^{-1}}(x+1))=\cos ({{\tan }^{-1}}x)\] \[\Rightarrow \]\[\frac{\pi }{2}-{{\cot }^{-1}}(x+1)=2n\pi \pm {{\tan }^{-1}}x\] Put\[n=0\Rightarrow \pi /2-{{\cot }^{-1}}(x+1)=\pm {{\tan }^{-1}}x={{\tan }^{-1}}(\pm x)\]\[\Rightarrow \]\[\pi /2={{\tan }^{-1}}(\pm x)+{{\cot }^{-1}}(x+1)\] \[\Rightarrow \]\[x+1=\pm x\Rightarrow 2x+1=0\]\[x=-\frac{1}{2}\]
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