A) \[1\]
B) \[-1\]
C) \[-1/2\]
D) x
Correct Answer: A
Solution :
Let \[u={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}\] ..?.(i) and \[\text{v}={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}\] ......(ii) In equation (i) put, \[x=\tan \theta \] \[\therefore \,\,u={{\tan }^{-1}}\left[ \frac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right]={{\tan }^{-1}}(\tan \,2\theta )\] \[\Rightarrow \,\,u=2\theta \,\,\Rightarrow \frac{du}{d\theta }=2\] ?..(a) In equation (ii), put \[x=\tan \theta \] \[\therefore \,\,\text{v=si}{{\text{n}}^{-1}}\left[ \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right]={{\sin }^{-1}}\,\,(\sin 2\theta )\] \[\Rightarrow \,\,\text{v = 2}\theta \Rightarrow \frac{dv}{d\theta }=2\] ?.. From equations and , \[\frac{du}{d\text{v}}=\frac{du}{d\theta }\times \frac{d\theta }{d\text{v}}=2\times \frac{1}{2}=1\] \[\therefore \] required differential coefficient will be 1.
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