A) \[a/b\]
B) \[ab\]
C) \[b/a\]
D) \[\frac{a-b}{1+ab}\]
Correct Answer: D
Solution :
[d] Given, \[{{\sin }^{-1}}\left( \frac{2a}{1+{{a}^{2}}} \right)-{{\cos }^{-1}}\left( \frac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)=ta{{b}^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\] \[\therefore 2{{\tan }^{-1}}a-2{{\tan }^{-1}}b=2{{\tan }^{-1}}x\] \[\Rightarrow {{\tan }^{-1}}a-{{\tan }^{-1}}b={{\tan }^{-1}}x\] \[\Rightarrow {{\tan }^{-1}}\left( \frac{a-b}{1+ab} \right)={{\tan }^{-1}}x\] \[\Rightarrow x=\frac{a-b}{1+ab}\]
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