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