A) \[-1\]
B) \[1\]
C) 2
D) 4
Correct Answer: B
Solution :
\[u={{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right),\upsilon =co{{s}^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] put \[x=\tan \theta \,x=\tan \theta \] \[u=2\theta \] \[u=2\theta \] \[\frac{du}{d\theta }=1\] \[\frac{du}{d\theta }=1\] \[\frac{du}{d\upsilon }=1\]You need to login to perform this action.
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