A) \[\frac{4}{1-{{x}^{2}}}\]
B) \[\frac{1}{1+{{x}^{2}}}\]
C) \[\frac{4}{1-{{x}^{2}}}\]
D) \[\frac{-4}{1+{{x}^{2}}}\]
Correct Answer: C
Solution :
Putting \[x=\tan \theta \] \[y={{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+{{\sec }^{-1}}\left( \frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } \right)\] \[=2\theta +2\theta =4{{\tan }^{-1}}x\]. \[y={{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+{{\sec }^{-1}}\left( \frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } \right)\] Þ \[x=0\].
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