\[\tan \left\{ \frac{1}{2}{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1}{2}{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right\}=\]

\[\tan \left\{ \frac{1}{2}{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1

1.	\[\tan \left\{ \frac{1}{2}{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1}{2}{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right\}=\]
A.	\[\frac{x-y}{1+xy}\]
B.	\[\frac{x+y}{1-xy}\]
C.	\[\frac{x-y}{x+y}\]
D.	\[\frac{1-xy}{1+xy}\]
Answer» C. \[\frac{x-y}{x+y}\]

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\[\tan \left\{ \frac{1}{2}{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+\frac{1}{2}{{\cos }^{-1}}\frac{1-{{y}^{2}}}{1+{{y}^{2}}} \right\}=\]

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