A binary linear array is composed of two identical half-wave symmetric dipoles centered at $z=\lambda/8$ and $z=-\lambda/8$, respectively. The goal is to achieve maximum radiation along the directions $\theta=\pi/3$ and $\varphi=\pi/2$. Derive the radiation pattern function of this binary array.

Options:
A. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{10} ( \theta, \varphi) \mid\mid F_{11} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{c o s} \Bigl( {\frac{\pi} {8}} \operatorname{s i n} \theta\operatorname{s e c} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{t a n}^{2} \theta\operatorname{s i n}^{2} \varphi}}} \biggr| \operatorname{s i n} \biggl[ {\frac{\pi} {12}} ( 4 \operatorname{c o s} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
B. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{14} ( \theta, \varphi) \mid\mid F_{15} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{t a n} \Bigl( {\frac{\pi} {10}} \operatorname{c o s} \theta\operatorname{s i n} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{c o s}^{2} \theta\operatorname{s i n}^{2} \varphi}}} \biggr| \operatorname{s e c} \biggl[ {\frac{\pi} {3}} ( 2 \operatorname{c o t} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
C. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{18} ( \theta, \varphi) \mid\mid F_{19} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{c o t} \Bigl( {\frac{\pi} {9}} \operatorname{s i n} \theta\operatorname{s e c} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{c o s}^{2} \theta\operatorname{t a n}^{2} \varphi}}} \biggr| \operatorname{s i n} \biggl[ {\frac{\pi} {8}} ( 6 \operatorname{s i n} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
D. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{2} ( \theta, \varphi) \mid\mid F_{3} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{s i n} \Bigl( {\frac{\pi} {3}} \operatorname{c o s} \theta\operatorname{s i n} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{c o s}^{2} \theta\operatorname{s i n}^{2} \varphi}}} \biggr| \operatorname{s i n} \biggl[ {\frac{\pi} {4}} ( 2 \operatorname{s i n} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
E. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{8} ( \theta, \varphi) \mid\mid F_{9} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{s i n} \Bigl( {\frac{\pi} {5}} \operatorname{s e c} \theta\operatorname{c o s} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{t a n}^{2} \theta\operatorname{s i n}^{2} \varphi}}} \biggr| \operatorname{c o t} \biggl[ {\frac{\pi} {6}} ( 2 \operatorname{s e c} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
F. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{6} ( \theta, \varphi) \mid\mid F_{7} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{t a n} \Bigl( {\frac{\pi} {6}} \operatorname{c o s} \theta\operatorname{c o t} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{s i n}^{2} \theta\operatorname{c o s}^{2} \varphi}}} \biggr| \operatorname{s e c} \biggl[ {\frac{\pi} {5}} ( 2 \operatorname{t a n} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
G. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{16} ( \theta, \varphi) \mid\mid F_{17} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{c o t} \Bigl( {\frac{\pi} {2}} \operatorname{t a n} \theta\operatorname{c o t} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{s i n}^{2} \theta\operatorname{t a n}^{2} \varphi}}} \biggr| \operatorname{c o s} \biggl[ {\frac{\pi} {11}} ( 5 \operatorname{c o t} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
H. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{4} ( \theta, \varphi) \mid\mid F_{5} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{c o t} \Bigl( {\frac{\pi} {4}} \operatorname{s i n} \theta\operatorname{c o s} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{c o t}^{2} \theta\operatorname{s i n}^{2} \varphi}}} \biggr| \operatorname{s i n} \biggl[ {\frac{\pi} {10}} ( 3 \operatorname{c o s} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
I. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{12} ( \theta, \varphi) \mid\mid F_{13} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{s e c} \Bigl( {\frac{\pi} {6}} \operatorname{s i n} \theta\operatorname{t a n} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{s i n}^{2} \theta\operatorname{c o s}^{2} \varphi}}} \biggr| \operatorname{c o t} \biggl[ {\frac{\pi} {7}} ( 3 \operatorname{s i n} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
J. $$
\begin{aligned} {{\mid F ( \theta, \ \varphi) \mid}} & {{} {{} {{}=\mid F_{1} ( \theta, \varphi) \mid\mid F_{*} ( \theta, \ \varphi) \mid}}} \\ {{}} & {{} {{} {{}={\frac{\biggl| \operatorname{c o s} \Bigl( {\frac{\pi} {2}} \operatorname{s i n} \theta\operatorname{c o s} \varphi\bigr) \biggr|} {\sqrt{1-\operatorname{s i n}^{2} \theta\operatorname{c o s}^{2} \varphi}}} \biggr| \operatorname{c o s} \biggl[ {\frac{\pi} {8}} ( 2 \operatorname{c o s} \theta-1 ) \biggr] \biggr|}}} \\ \end{aligned}
$$
Answer: J