A) ?1
B) 1
C) 2
D) 4
Correct Answer: B
Solution :
Let \[p={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}=2{{\tan }^{-1}}x\] and \[q={{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}=2{{\tan }^{-1}}x\]; \ \[\frac{dp}{dq}=\frac{dp/dx}{dq/dx}=1\].You need to login to perform this action.
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