A) \[\frac{1}{2}\cos 2A\]
B) \[\frac{1}{2}\sin 2A\]
C) \[\frac{1}{2}\tan 2A\]
D) \[\cot 2A\]
Correct Answer: B
Solution :
\[{{\sin }^{2}}(15{}^\circ +A)-{{\sin }^{2}}(15{}^\circ -A)\] \[=\sin \,\,(15{}^\circ +A+15{}^\circ -A)\cdot sin\,\,(15{}^\circ +A-15{}^\circ +A)\] \[[\because {{\sin }^{2}}A-{{\sin }^{2}}B=\sin \,\,(A+B)\cdot \sin \,\,(A-B)]\] \[=\sin 30{}^\circ \sin \,\,(2A)=\frac{1}{2}\sin 2A\]
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