Respuesta :
[tex]\tan^3\theta-\dfrac1{\tan\theta-1}-\sec^2\theta+1=0[/tex]
Recall that [tex]\tan^2\theta+1=\sec^2\theta[/tex], so we can write everything in terms of [tex]\tan\theta[/tex]:
[tex]\tan^3\theta-\dfrac1{\tan\theta-1}-\tan^2\theta=0[/tex]
Let [tex]x=\tan\theta[/tex], so that
[tex]x^3-\dfrac1{x-1}-x^2=0[/tex]
With some rewriting we get
[tex]x^3-x^2-\dfrac1{x-1}=0[/tex]
[tex]x^2(x-1)-\dfrac1{x-1}=0[/tex]
[tex]\dfrac{x^2(x-1)^2-1}{x-1}=0[/tex]
Clearly we cannot have [tex]x=1[/tex], or [tex]\tan\theta=1[/tex].
The numerator determines when the expression on the left reduces to 0:
[tex]x^2(x-1)^2-1=0[/tex]
[tex]x^2(x-1)^2=1[/tex]
[tex]\sqrt{x^2(x-1)^2}=\sqrt1[/tex]
[tex]|x(x-1)|=1[/tex]
[tex]x(x-1)=1\text{ or }x(x-1)=-1[/tex]
Completing the square gives
[tex]x(x-1)=x^2-x=x^2-x+\dfrac14-\dfrac14=\left(x-\dfrac12\right)^2-\dfrac14[/tex]
so that
[tex]\left(x-\dfrac12\right)^2=\dfrac54\text{ or }\left(x-\dfrac12\right)^2=-\dfrac34[/tex]
The second equation gives no real-valued solutions because squaring any real number gives a positive real number. (I'm assuming we don't care about complex solutions.) So we're left with only
[tex]\left(x-\dfrac12\right)^2=\dfrac54[/tex]
[tex]\sqrt{\left(x-\dfrac12\right)^2}=\sqrt{\dfrac54}[/tex]
[tex]\left|x-\dfrac12\right|=\dfrac{\sqrt5}2[/tex]
which again gives two cases,
[tex]x-\dfrac12=\dfrac{\sqrt5}2\text{ or }x-\dfrac12=-\dfrac{\sqrt5}2[/tex]
[tex]x=\dfrac{1+\sqrt5}2\text{ or }x=\dfrac{1-\sqrt5}2[/tex]
Then when [tex]x=\tan\theta[/tex], we can find [tex]\cot\theta[/tex] by taking the reciprocal, so we get
[tex]\boxed{\cot\theta=\dfrac2{1+\sqrt5}\text{ or }\cot\theta=\dfrac2{1-\sqrt5}}[/tex]
