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In acute angled $\Delta ABC$ prove that,

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

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, $\sin B > 1 - \cos B$ & $\sin C > 1 - \cos C$

On addition, $\sin A + \sin B + \sin C > 3 - (\cos A + \cos B + \cos C)$ ........(!)

In any triangle, $\cos A + \cos B + \cos C \le \frac{3}{2}$*

$\therefore 2(\cos A + \cos B + \cos C) \le 3$

$ \Rightarrow \cos A + \cos B + \cos C \le 3 - (\cos A + \cos B + \cos C)$ ........(!!)

From (!) & (!!), $\sin A + \sin B + \sin C > \cos A + \cos B + \cos C$

*Please refer textbook or study material for the proof.

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