Question

If $0<\alpha <\beta <\gamma <\pi /2$, then show $\tan \alpha <\frac{\sin \alpha +\sin \beta +\sin \gamma }{\cos \alpha +\cos \beta +\cos \gamma }<\tan \gamma$

Answer 1

$\because$ $0<\alpha <\beta <\gamma <\frac{\pi }{2}$
$\therefore$ $\sin \alpha <\sin \beta <\sin \gamma$
and $\cos \alpha >\cos \beta >\cos \gamma$

So $\frac{\sin \alpha +\sin \beta +\sin \gamma }{\cos \alpha +\cos \beta +\cos \gamma }<\frac{3\sin \gamma }{3\cos \gamma }>\frac{3\sin \alpha }{3\cos \alpha }$

Thus, $\tan \alpha <\frac{\sin \alpha +\sin \beta +\sin \gamma }{\cos \alpha +\cos \beta +\cos \gamma }<\tan \gamma$