A) 1
B) \[\frac{3}{2}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
Let \[{{y}_{1}}={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\,\,\,=2{{\tan }^{-1}}x\], \[{{\cot }^{-1}}\left( \frac{1-3{{x}^{2}}}{3x-{{x}^{3}}} \right)=3{{\tan }^{-1}}x\Rightarrow \frac{d{{y}_{1}}}{d{{y}_{2}}}=\frac{\left( \frac{d{{y}_{1}}}{dx} \right)}{\left( \frac{d{{y}_{2}}}{dx} \right)}=\frac{\left( \frac{2}{1+{{x}^{2}}} \right)}{\left( \frac{3}{1+{{x}^{2}}} \right)}=\frac{2}{3}\]
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