Q9:suppose that 0

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Q9:suppose that 0

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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not sure how to do this, but I would use a comparison test.. maybe showing that first: tan xn = sin xn/cos xn = nsinxn / ncosxn then showing: [sin x is increasing on x e (0, pi/2) ] (n-1)(sin x(n-1)) + sin xn < nsinxn [cos x is decreasing on x e (0, pi/2) ] and (n-1)(cos x(n-1)) + cos xn > ncosxn Then [(n-1)(sin x(n-1)) + sin xn]/[(n-1)(cos x(n-1)) + cos xn] < n sin xn/ n cos xn Similar reasoning to show it's > tan x1. but ye, Just a suggestion, it's probably wrong. but ye use if you can..
what is a (n-1)sinx(n-1)?and how you use that?

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it's \[\sin x_{n-1}\] n-1 times... so \[\sin x_{n-1} +\sin x_{n-1} +\sin x_{n-1} +\sin x_{n-1} + ... +\sin x_{n-1}\]
I know !bot I want to solve: tanx1<(sinx1+sinx2+...+sinxn)/(cosx1+cosx2+...+cosxn)
well, you can't solve that. I'm saying you can use a comparison to prove it... Read my first message thoroughly
I see your first commend. But choosing those two terms in num and den confusing. However I accept your opinion for sin and cos.
\[[n.\sin x _{1} \le \sin x _{1}+\sin x _{2}+...\sin x _{n}\le n.\sin x _{n}\\] n.\cos x _{n} \le \cos x _{1}+\cos x _{2}+...\cos x _{n}\le n.\cos x _{1}\]\] as a known a a/C

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