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

Question

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.

Answer 1

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.

Answer 2

$\tan x$ 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.

Answer 3

$\tan x - \cot x =2$

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

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

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

$-\cos 2x=\sin2x$

$\tan 2x=-1$

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

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