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}$

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

In acute angled triangle, $\cos \frac{A}{2} > \sin \frac{A}{2}$

So, $\sin \frac{A}{2}.\cos \frac{A}{2} > \sin \frac{A}{2}.\sin \frac{A}{2}$

Hence, $2\sin \frac{A}{2}\cos \frac{A}{2} > 2{\sin ^2}\frac{A}{2}$

Thus, $\sin A > 1 – \cos A$

Likewise, $\sin B > 1 – \cos B$ & $\sin C > 1 – \cos C$ Continue reading In acute angled $\Delta ABC$ prove that,

$sin A + sin B + sin C > cos A + cos B + cos C$ →