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Trignometry

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15 Jan 2009 14:52:43 IST

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Solution of Quadratic Trig System

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This problem I have framed on my own and I am a novice at that. So please pardon me if its too easy or bugging!Solve the following system in for the case n is odd:\sec^2 \theta_2 + \sec \theta_2 \sec \theta_3 + \sec^2 \theta_3 = 1 \\ \\. \\ \\. \\ \\. \\ \\\sec^2 \theta_{n-1} + \sec \theta_{n-1} \sec \theta_n + \sec^2 \theta_n = 1 \\ \\\sec^2 \theta_n + \sec \theta_n \sec \theta_1 + \sec^2 \theta_1 = 1" src="http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/2/9/7/297dc4fc8a6e47babf45a7a3134d08779e176a2d.gif" align="absMiddle" minmax_bound="true" />

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Comments (5)

Hari Shankar

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16 Jan 2009 19:19:05 IST

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First time any of my problems have lasted this long. Maybe I should have sent it to the JEE Coordination team instead

Arnab Chowdhuri

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17 Jan 2009 15:24:32 IST

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we know that sec (*)<=-1 or sec(*)>=1 hence sec(*)^2 is always greater than 1.hence the given system of equation cannt have any solution for theta

Hari Shankar

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17 Jan 2009 18:46:28 IST

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no, not right.

Arnab Chowdhuri

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19 Jan 2009 12:02:39 IST

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Iam sorry sir. Please post a solution.

Hari Shankar

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19 Jan 2009 12:22:25 IST

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The idea is very straighforward.

Since $\sec^2 \theta_1 + \sec^2 \theta_2 \ge 2$ we must have $\sec \theta_1 \sec \theta_2 < 0$

So, if we start with supposing that $\sec \theta_1 > 0$ , we must have $\sec \theta_2<0$ which in turn implies $\sec \theta_3>0$ and so on till we obtain in the case when n is odd, $\sec \theta_1<0$

A similar contradiction occurs if we start with the assumption that $\sec \theta_1<0$ . So no solutions exist.

Does this explain the big grin in my question?

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