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It's been 20 years since I did trig, and this one seems a little tricky. How would I solve $$ \tan^2(x) -2\tan(x)=1 $$ with steps?

Anton Grudkin
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MaLio
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    Let $\tan x = t,$ and solve $t^2-2t=1,$ first. – dxiv Apr 01 '17 at 18:34
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    @MaLio. You will need to quadratic formula. Still remember that one? – imranfat Apr 01 '17 at 18:42
  • yes, however I don't think that's the way to go about solving the problem. This is in my daughters textbook in the trig identities section. (yes, I am trying to help with homework). I like haqnatural solution below, but how does the tan^2x-2tanx-> (tanx -1)^2 - 1 ? – MaLio Apr 01 '17 at 18:46
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    If she is learning trig identities, then she likely knows the quadratic formula and can use it. – Kaynex Apr 01 '17 at 18:58

3 Answers3

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$$\tan ^{ 2 }{ x-2\tan { x } -1=0 } \\ \tan ^{ 2 }{ x-2\tan { x } +1-2=0 } \\ { \left( \tan { x } -1 \right) }^{ 2 }-2=0\\ \tan { x-1=\pm \sqrt { 2 } } \\ \tan { x } =1\pm \sqrt { 2 } \\ x=\arctan { \left( 1\pm \sqrt { 2 } \right) +\pi n } \\ $$

haqnatural
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let $B = \tan(x)$

$\tan(x)$ varies with $x$, but ultimately is just a value

now rewriting the equation to give $B^2 -2B -1 = 0$, This will not factorise with integers, but solves to give $x = 1$ plus or minus square root $2$

from there, we use the $\arctan$ function and it tells you that (in radians) x = $\arctan(1 + \sqrt2)$ or $\arctan(1 - \sqrt2$) which will give you two roots every $2\pi$ radians, so you need to restrict the range of the function to get any real answers

Sri-Amirthan Theivendran
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Alex Robinson
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Put $\tan x = z$

Then equation becomes,

$z^2 - 2z = 1$

$z^2 - 2z - 1 = 0$

Hope you can now factorise.

Kanwaljit Singh
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