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Question

how to evaluate $\tan x-\cot x=2.$

Given that it lies between on $\left[\frac{-\pi} 2,\frac \pi 2 \right]$.

My Steps so far

I converted cot into tan to devolve into $\frac{\tan^2 x-1}{\tan x}=2$.

Then I multiply $\tan{x}$ on both sides and then get $\tan^2 x-2\tan x-1$.

From there I dont know where to go.

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    $\begingroup$ Let $u = \text{tan}(x)$. Solve the new equation for $u$. $\endgroup$
    – quasi
    Jan 21, 2017 at 5:13
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    $\begingroup$ Often people here write "Solve" when "Evalutate" is appropriate, but here you've got "Evaluate" where "Solve" is appropriate. $\endgroup$ Jan 21, 2017 at 5:13
  • $\begingroup$ Also, your sentence "Given that it lies ..." would be clearer if written as "Given that $x$ lies ...". $\endgroup$
    – quasi
    Jan 21, 2017 at 5:15
  • $\begingroup$ @quasi that is too frivolous. I think that the spirit of the post is what matters. Please stay on topic of math not grammar. and I think that U sub is useless in this scenario, please show me how you did it $\endgroup$
    – John Rawls
    Jan 21, 2017 at 5:17
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    $\begingroup$ "$u$ sub" is not useless in this scenario. Please refrain from being mean to people who are trying to help you $\endgroup$
    – user223391
    Jan 21, 2017 at 5:18

3 Answers 3

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$\tan x - \cot x =2$

$\dfrac{\sin x }{\cos x} - \dfrac{\cos x}{\sin x }=2$

$\dfrac{\sin^2 x - \cos ^2 x}{\sin x \cos x } =2$

$-\cos 2x = 2\sin x \cos x$

$-\cos 2x=\sin2x$

$\tan 2x=-1$

$2x= n\pi -\frac{\pi}{4}$

Put n values to get values of x in required range .

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$\tan x$ and $\cot x$ are each other's reciprocals. Let $u=\tan x,$ so that $\dfrac 1 u = \cot x.$

Then $\tan x - \cot x = 2$ becomes $u - \dfrac 1 u = 2,$ and multiplying both sides by $u$ yields $u^2 - 1 = 2u,$ a quadratic equation. You get $u=1\pm\sqrt2.$

Next, if $\tan x= 1+\sqrt2,$ then what is $x$? Perhaps just expressing this as a value of the arctangent function suffices.

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    $\begingroup$ (+1) Your answer was the first posted. It has far more information content than the "accepted answer," and yet was pushed down the stack. I continue to be bewildered by voting behavior herein. Oh and Happy New Year Michael! -Mark $\endgroup$
    – Mark Viola
    Jan 21, 2017 at 6:06
  • $\begingroup$ @Dr.MV. Hi, Mark ! Happy New Year ! I fully agree with your comment. $\endgroup$ Jan 21, 2017 at 6:18
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We then can solve the quadratic equation thus getting $$\tan x =\frac {2 \pm \sqrt{4-4 (1)(-1)}}{2} = \frac {2\pm 2\sqrt {2}}{2} = 1\pm \sqrt {2} $$ Thus $$x =\arctan ( 1\pm \sqrt {2}) $$ Hope it helps.

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