- A$\sin (\alpha + \beta )\sin (\alpha - \beta )$
- B$\cos (\alpha + \beta )\cos (\alpha - \beta )$
- C$\sin (\alpha - \beta )\cos (\alpha + \beta )$
- ✓$\sin (\alpha + \beta )\cos (\alpha - \beta )$
$ = \cos \,\left( {\frac{\pi }{4} - \beta + \alpha - \frac{\pi }{4}} \right)\,\cos \,\left( {\frac{\pi }{4} - \beta - \alpha + \frac{\pi }{4}} \right)\,$
$ = \cos (\alpha - \beta )\cos \left( {\frac{\pi }{2} - \overline {\alpha + \beta } } \right) $
$= \cos (\alpha - \beta )\sin (\alpha + \beta )$.
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$ A=\{z: \operatorname{Im} z \geq 1\} $
$ B=\{z:|z-2-i|=3\} $
$ C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\} .$
$1.$ The number of elements in the set $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}$ is
$(A)$ $0$ $(B)$ $1$ $(C)$ $2$ $(D)$ $\infty$
$2.$ Let $z$ be any point in $A \cap B \cap C$. Then, $|z+1-i|^2+|z-5-i|^2$ lies between
$(A)$ $25$ and $29$ $(B)$ $30$ and $34$ $(C)$ $35$ and $39$ $(D)$ $40$ and $44$
$3.$ Let $z$ be any point in $A \cap B \cap C$ and let $w$ be any point satisfying $|w-2-i|<3$. Then, $|z|-|w|+3$ lies between
$(A)$ $-6$ and $3$ $(B)$ $-3$ and $6$
$(C)$ $-6$ and $6$ $(D)$ $-3$ and $9$
Give the answer question $1,2$ and $3.$