A) \[\sqrt{\frac{1+{{x}^{2}}}{2+{{x}^{2}}}}\]
B) \[\sqrt{\frac{2+{{x}^{2}}}{2+{{x}^{2}}}}\]
C) \[\sqrt{\frac{1+{{x}^{2}}}{2-{{x}^{2}}}}\]
D) \[\sqrt{\frac{1-{{x}^{2}}}{2+{{x}^{2}}}}\]
Correct Answer: A
Solution :
[a] \[\because sin[co{{t}^{-1}}\{cos(ta{{n}^{-1}}x)\}]\] \[=\sin \left\{ {{\cot }^{-1}}\left( \cos \left( {{\cos }^{-1}}\frac{1}{\sqrt{1+{{x}^{2}}}} \right) \right) \right\}\] \[=\sin \left( {{\cot }^{-1}} \right)\left( \frac{1}{\sqrt{1+{{x}^{2}}}} \right)=\sin \left( {{\sin }^{-1}}\frac{\sqrt{1+{{x}^{2}}}}{\sqrt{{{x}^{2}}+2}} \right)\] \[=\frac{\sqrt{1+{{x}^{2}}}}{\sqrt{{{x}^{2}}+2}}=\sqrt{\frac{1+{{x}^{2}}}{2+{{x}^{2}}}}\] Hence, option [a] is correct.
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