MCQ
If $\sin A = \sin B$ and $\cos A = \cos B,$ then
- ✓$\sin \frac{{A - B}}{2} = 0$
- B$\sin \frac{{A + B}}{2} = 0$
- C$\cos \frac{{A - B}}{2} = 0$
- D$\cos (A + B) = 0$
$\frac{{\sin A}}{{\sin B}} = \frac{{\cos A}}{{\cos B}}\,$
$ \Rightarrow \,\,\sin A\,\cos B - \cos A\,\sin B = 0$
$ \Rightarrow \,\,\sin \,(A - B) = 0$
Hence, $\sin \,\left( {\frac{{A - B}}{2}} \right) = 0.$
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$\mathrm{S}_{1}=\{\mathrm{z} \in \mathrm{C}:|\mathrm{z}-2| \leq 1\} \text { and }$
$\mathrm{S}_{2}=\{\mathrm{z} \in \mathrm{C}: \mathrm{z}(1+\mathrm{i})+\overline{\mathrm{z}}(1-\mathrm{i}) \geq 4\}$
Then, the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in \mathrm{S}_{1} \cap \mathrm{S}_{2}$ is equal to:
$(A)\,\,\,5\%$ families own both a car and a phone
$(B)\,\,\,35\%$ families own either a car or a phone
$(C)\,\,\,40,000$ families live in the town
Then,