If \[\sin x+\sin y=a\] and \[cos\text{ }x+cos\,y=b,\]then \[{{\tan }^{2}}\left( \frac{x+y}{2} \right)+{{\tan }^{2}}\left( \frac{x-y}{2} \right)\] is equal to

\[\frac{{{a}^{4}}-{{b}^{4}}+4{{a}^{2}}}{{{a}^{2}}{{b}^{2}}+{{a}^{4}}}\]

\[\frac{{{a}^{4}}+{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]

\[\frac{{{a}^{4}}-{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]

If \[\sin x+\sin y=a\] and \[cos\text{ }x+cos\,y=b,\]then \[{{\tan }^{

1.	If \[\sin x+\sin y=a\] and \[cos\text{ }x+cos\,y=b,\]then \[{{\tan }^{2}}\left( \frac{x+y}{2} \right)+{{\tan }^{2}}\left( \frac{x-y}{2} \right)\] is equal to
A.	\[\frac{{{a}^{4}}+{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]
B.	\[\frac{{{a}^{4}}-{{b}^{4}}+4{{b}^{2}}}{{{a}^{2}}{{b}^{2}}+{{b}^{4}}}\]
C.	\[\frac{{{a}^{4}}-{{b}^{4}}+4{{a}^{2}}}{{{a}^{2}}{{b}^{2}}+{{a}^{4}}}\]
D.	None of the above
Answer» C. \[\frac{{{a}^{4}}-{{b}^{4}}+4{{a}^{2}}}{{{a}^{2}}{{b}^{2}}+{{a}^{4}}}\]

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If \[\sin x+\sin y=a\] and \[cos\text{ }x+cos\,y=b,\]then \[{{\tan }^{2}}\left( \frac{x+y}{2} \right)+{{\tan }^{2}}\left( \frac{x-y}{2} \right)\] is equal to

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