A) \[\sqrt{\frac{({{x}^{2}}+1)}{({{x}^{2}}+2)}}\]
B) \[\sqrt{\frac{({{x}^{2}}+2)}{({{x}^{2}}+1)}}\]
C) \[\frac{({{x}^{2}}+1)}{({{x}^{2}}+2)}\]
D) \[\frac{({{x}^{2}}+2)}{({{x}^{2}}+1)}\]
Correct Answer: A
Solution :
[a] Let \[\alpha ={{\tan }^{-1}}x\Rightarrow \tan \alpha =x\] then \[\cos \alpha =\frac{1}{\sqrt{1+{{\tan }^{2}}\alpha }}=\frac{1}{\sqrt{1+{{x}^{2}}}}\] \[\Rightarrow \cos (ta{{n}^{-1}}x)\}=\left\{ \frac{1}{\sqrt{1+{{x}^{2}}}} \right\}\] So, \[{{\cot }^{-1}}\cos (ta{{n}^{-1}}x)=co{{t}^{-1}}\left\{ \frac{1}{\sqrt{1+{{x}^{2}}}} \right\}\] Let \[{{\cot }^{-1}}\left( \frac{1}{\sqrt{1+{{x}^{2}}}} \right)=\beta \] \[\Rightarrow \cot \beta =\frac{1}{\sqrt{1+{{x}^{2}}}}\] and \[\sin \beta =\frac{1}{\sqrt{1+{{\cot }^{2}}\beta }}\] \[=\frac{\sqrt{1+{{x}^{2}}}}{\sqrt{{{x}^{2}}+1+1}}=\sqrt{\frac{{{x}^{2}}+1}{{{x}^{2}}+2}}\] \[\Rightarrow \sin [co{{t}^{-1}}\{cos(ta{{n}^{-1}})\}]=\sqrt{\frac{{{x}^{2}}+1}{{{x}^{2}}+2}}\]
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