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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?

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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!

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  • $\begingroup$ I don't really understand $\endgroup$ – Pragna Aug 13 '17 at 5:07
  • $\begingroup$ I am in 11th standard and this was given in our exam. Noone really solved. It will be very helpful if you explain it $\endgroup$ – Pragna Aug 13 '17 at 5:08
  • $\begingroup$ @Pragna I used $\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$. $\endgroup$ – Michael Rozenberg Aug 13 '17 at 5:09
  • $\begingroup$ Ok yes... I understand $\endgroup$ – Pragna Aug 13 '17 at 5:10
  • $\begingroup$ @ Pragna You are welcome! $\endgroup$ – Michael Rozenberg Aug 13 '17 at 5:11

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