A) \[x-y\]
B) \[x+y\]
C) \[\frac{x-y}{1+xy}\]
D) \[\frac{x+y}{1-xy}\]
Correct Answer: C
Solution :
Given that \[{{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}A\] \[\Rightarrow \,\,{{\tan }^{-1}}\,\left( \frac{x-y}{1+xy} \right)={{\tan }^{-1}}A\]. Hence\[A=\frac{x-y}{1+xy}\].
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