A) \[\frac{3}{2}\]
B) 1
C) \[\frac{1}{2}\]
D) \[\frac{2}{3}\]
Correct Answer: D
Solution :
Let \[u={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right),\,\upsilon ={{\cos }^{-1}}\left( \frac{1-3{{x}^{2}}}{3x-{{x}^{3}}} \right)\] Put \[x=\tan \theta \] \[\therefore \] \[u={{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\] \[u={{\cos }^{-1}}\cos 2\theta \left[ \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }=\cos 2\theta \right]\] \[u=2\theta \] \[\therefore \] \[\frac{du}{d\theta }=2\] and \[\upsilon ={{\cot }^{-1}}\left( \frac{1-3{{\tan }^{2}}\theta }{3\tan \theta -{{\tan }^{3}}\theta } \right)\] \[={{\cot }^{-1}}\cot 3\theta \] \[\left[ [\cot 3\theta =\left( \frac{1-3{{\tan }^{2}}\theta }{3\tan \theta -{{\tan }^{3}}\theta } \right) \right]\] \[\upsilon =3\theta \] \[\frac{d\upsilon }{d\theta }=3\] \[\therefore \] \[\frac{du}{d\upsilon }=\frac{du}{d\theta }\times \frac{d\theta }{d\upsilon }\] \[=2\times \frac{1}{3}=\frac{2}{3}\]
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