\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}\]is equal to

\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}\]is

1.	\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}\]is equal to
A.	\[\tan (A-B)\]
B.	\[\tan (A+B)\]
C.	\[cot(A-B)\]
D.	\[cot(A+B)\]
Answer» C. \[cot(A-B)\]

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\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}\]is equal to

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