A) \[\tan (A-B)\]
B) \[\tan (A+B)\]
C) \[\cot (A-B)\]
D) \[\cot (A+B)\]
Correct Answer: B
Solution :
\[\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{\sin A\cos A-\sin B\cos B}=\frac{2\,\sin \,(A+B)\,\sin \,(A-B)}{\sin \,2A-\sin \,2B}\] \[=\frac{2\,\sin \,(A+B)\,\sin \,(A-B)}{2\,\cos \,(A+B)\,\sin \,(A-B)}=\tan \,(A+B)\].
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