The given expression can be written as,

$({a^2}{\sec ^2}\theta – {a^2}) + {a^2} + ({b^2}{\rm{cose}}{{\rm{c}}^2}\theta – {b^2}) + {b^2}$

$ = {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta + {a^2} + {b^2}$

Using A.M. $\geq $ G.M.,

$\frac{{{a^2}{{\tan }^2}\theta + {b^2}{{\cot }^2}\theta }}{2} \ge \sqrt {{a^2}{{\tan }^2}\theta .{b^2}{{\cot }^2}\theta } $

$ \Rightarrow {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta \ge 2|ab|$

$ \Rightarrow {a^2}{\tan ^2}\theta + {b^2}{\cot ^2}\theta + {a^2} + {b^2} \ge 2|ab| + {a^2} + {b^2}$

$\therefore{({a^2}{\sec ^2}\theta + {b^2}{\rm{cose}}{{\rm{c}}^2}\theta )_{\min }} = {a^2} + {b^2} + 2|ab|$