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solve this equation :

$(2-\sec^21)(2-\sec^22)(2-\sec^23)........(2-\sec^288)(2-\sec^289)$

If tried from tangent approach with$(1+1-sec^21)......(1+1-\sec^289)$

and i do (1,89) ; (2,88);and........ so on so i get

$(1-\tan^21)(1-\tan^289)$

$(1+\tan^21\tan^289-\tan^289-\tan^21)$

i've got $\tan^21\tan^289= 1 $

from $\tan(89+1)=\frac{\tan89+\tan1}{1-\tan89\tan1}$

but i don't know how to get $(\tan^289+\tan^21)$

and i don't know how to continue it

and please from other approach too

thank you

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2 Answers 2

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You get $$\prod_{0\le x\le 89}(1-\tan^2x^\circ)\text{ not }\prod_{0\le x\le 89}(1+\tan^2x^\circ)$$

But $\tan45^\circ=1$

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    $\begingroup$ Yes, so the answer is 0 i'm very stupid Thank you very much it's so much helping $\endgroup$
    – freeze
    Apr 17, 2013 at 8:27
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Hint: What is $\cos 45^{\circ}$?

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  • $\begingroup$ $\frac{2^{\frac{1}{2}}}{2}$ but what is the relationship? $\endgroup$
    – freeze
    Apr 17, 2013 at 8:18
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    $\begingroup$ @freeze: I suggest you don't just stop there, and see how you might use that hint. Here is another puzzle of the same kind: what is $(x-a)(x-b)(x-c)\dots(x-z)?$ $\endgroup$
    – Aryabhata
    Apr 17, 2013 at 8:21
  • $\begingroup$ but i don't know how to get $(\tan^289+tan^21)$? can you help me $\endgroup$
    – freeze
    Apr 17, 2013 at 8:24
  • $\begingroup$ @freeze: What's the value of this : $(4-1)(4-2)(4-3)(4-4)(4-5)..(4-n)$? $\endgroup$
    – Inceptio
    Apr 17, 2013 at 9:34

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