Question

If tan^3(theta) -1/tan(theta)-1 - sec^2(theta) + 1 = 0 find cot(theta)

Answer 1

$\tan^3\theta-\dfrac1{\tan\theta-1}-\sec^2\theta+1=0$

Recall that $\tan^2\theta+1=\sec^2\theta$, so we can write everything in terms of $\tan\theta$:

$\tan^3\theta-\dfrac1{\tan\theta-1}-\tan^2\theta=0$

Let $x=\tan\theta$, so that

$x^3-\dfrac1{x-1}-x^2=0$

With some rewriting we get

$x^3-x^2-\dfrac1{x-1}=0$

$x^2(x-1)-\dfrac1{x-1}=0$

$\dfrac{x^2(x-1)^2-1}{x-1}=0$

Clearly we cannot have $x=1$, or $\tan\theta=1$.

The numerator determines when the expression on the left reduces to 0:

$x^2(x-1)^2-1=0$

$x^2(x-1)^2=1$

$\sqrt{x^2(x-1)^2}=\sqrt1$

$|x(x-1)|=1$

$x(x-1)=1\text{ or }x(x-1)=-1$

Completing the square gives

$x(x-1)=x^2-x=x^2-x+\dfrac14-\dfrac14=\left(x-\dfrac12\right)^2-\dfrac14$

so that

$\left(x-\dfrac12\right)^2=\dfrac54\text{ or }\left(x-\dfrac12\right)^2=-\dfrac34$

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

$\left(x-\dfrac12\right)^2=\dfrac54$

$\sqrt{\left(x-\dfrac12\right)^2}=\sqrt{\dfrac54}$

$\left|x-\dfrac12\right|=\dfrac{\sqrt5}2$

which again gives two cases,

$x-\dfrac12=\dfrac{\sqrt5}2\text{ or }x-\dfrac12=-\dfrac{\sqrt5}2$

$x=\dfrac{1+\sqrt5}2\text{ or }x=\dfrac{1-\sqrt5}2$

Then when $x=\tan\theta$, we can find $\cot\theta$ by taking the reciprocal, so we get

$\boxed{\cot\theta=\dfrac2{1+\sqrt5}\text{ or }\cot\theta=\dfrac2{1-\sqrt5}}$