\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}=\] [MP PET 1993]

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

1.	\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}=\] [MP PET 1993]
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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\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}=\] [MP PET 1993]

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