A) \[\frac{a-b}{1+ab}\]
B) \[\frac{b}{1+ab}\]
C) \[\frac{b}{1-ab}\]
D) \[\frac{a+b}{1-ab}\]
Correct Answer: D
Solution :
\[{{\sin }^{-1}}\left( \frac{2a}{1+{{a}^{2}}} \right)+{{\sin }^{-1}}\left( \frac{2b}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x\] Putting \[a=\tan \theta \]and \[b=\tan \varphi \] So, \[{{\sin }^{-1}}\left( \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+{{\sin }^{-1}}\left( \frac{2\tan \varphi }{1+{{\tan }^{2}}\varphi } \right)=2{{\tan }^{-1}}x\] Þ \[{{\sin }^{-1}}\sin (2\theta )+{{\sin }^{-1}}\sin (2\varphi )=2{{\tan }^{-1}}x\] Þ \[2(\theta +\varphi )=2{{\tan }^{-1}}x\] Hence \[x=\tan (\theta +\varphi )\]Þ \[x=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }\] Substituting these values, we get \[x=\frac{a+b}{1-ab}\].
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