A) \[x=\frac{a-b}{1+ab}\]
B) \[x=\frac{a+b}{1-ab}\]
C) \[x=\frac{{{a}^{2}}-{{b}^{2}}}{1+ab}\]
D) None of these
Correct Answer: B
Solution :
[b] \[\because si{{n}^{-1}}\frac{2a}{1+{{a}^{2}}}+si{{n}^{-1}}\frac{2b}{1+{{b}^{2}}}=2ta{{n}^{-1}}x\] \[\Rightarrow 2ta{{n}^{-1}}a+2ta{{n}^{-1}}b=2ta{{n}^{-1}}x\] \[\left[ \because \frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta \right]\] \[\Rightarrow ta{{n}^{-1}}a+ta{{n}^{-1}}b=ta{{n}^{-1}}x\] \[\Rightarrow ta{{n}^{-1}}\left[ \frac{a+b}{1-ab} \right]=ta{{n}^{-1}}x\] \[\Rightarrow x=\frac{a+b}{1-ab}\] Hence, option [b] is correct.
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