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How can I assign $a,b,c,d$ values $\pm \tan\theta,\pm{1\over\tan\theta}$ so that ${(a-b)(c-d)\over (a-d)(b-c)}=\tan^2(2\theta)$? Thank you for helping.

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    @Sp3000: actually the other tan related factor is a reciprocal, also I think the 2's are unnecessary... I have edited the question, hope it makes more sense now.2012-02-08

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Since , $\tan^2 2\theta =\frac{4\tan^2\theta}{(1-\tan^2\theta)^2}$ solution is :

$a=\frac{-1}{\tan \theta}$ , $b=\frac{1}{\tan \theta}$ , $c=\tan \theta$ , $d=-\tan \theta$

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    @DominicD,In that case I don't think that is possible to express $\tan^2 2\theta$...2012-02-08