If $0<\alpha_1<\alpha_2<\cdots<\alpha_n<\frac{\pi}{2}$ is given then prove: $$\tan\alpha_1 < \frac{\sin\alpha_1+\sin\alpha_2+\sin\alpha_3+\cdots+\sin\alpha_n}{\cos\alpha_1+\cos\alpha_2+\cos\alpha_3+\cdots+\cos\alpha_n} < \tan\alpha_n$$
How should I solve this?
Te left inequality.
We need to prove that $$\frac{\sin\alpha_1}{\cos\alpha_1}<\frac{\sin\alpha_1+\cdots+\sin{\alpha_n}}{\cos\alpha_1+\cdots+\cos\alpha_n}$$ or $$\sin(\alpha_1-\alpha_2)+\cdots+\sin(\alpha_1-\alpha_n)<0,$$ which is obvious.
We can prove the right inequality by the same way.
Finally we'll get there $$\sin(\alpha_n-\alpha_1)+\cdots+\sin(\alpha_n-\alpha_{n-1})>0.$$ Done!