# Elementary trigonometry: $\tan$

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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I'm not exactly sure how this works. If you assign values $\alpha \tan \theta$, $\beta \tan \theta ...$ to $a, b, c, d$ respectively, where $\alpha, \beta ...$ correspond to one of $\pm 2, \pm \frac{1}{2}$ then you can cancel out the $\tan \theta$s and you'll be left with a constant. –  Sp3000 Feb 8 '12 at 9:42
@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. –  Dominic D Feb 8 '12 at 10:11

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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Oh it does check out (and the 2s really were unnecessary). Out of curiosity may I ask how you got that answer or did you just try every possibility? –  Sp3000 Feb 8 '12 at 10:58
@Sp3000,I had a moment of lucidity :-) –  pedja Feb 8 '12 at 11:04
Thanks, pedja. I have removed the 2's in the question, is it possible to get an answer if the 2's are still in? I am asking this because They are actually in the question on my homework sheet... –  Dominic D Feb 8 '12 at 11:24
Personally i don't think it is possible if the 2's are in because you cant get the bottom bit of the fraction as $1-\tan^2\theta$ there must be a factor of 4 in front , right? –  Dominic D Feb 8 '12 at 11:35
I meant a factor of 4 in front of $\tan\theta$. I am unable to find a combination to get $(\tan^2(2\theta))/2$ would you mind shedding some light? –  Dominic D Feb 8 '12 at 11:46