SOLUTION: Please help me solve this problem: If {{{ sqrt(3)cot^2 - 4cot + sqrt(3)= 0 }}} Then find cot^2 + tan^2

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Question 789080: Please help me solve this problem:
If +sqrt%283%29cot%5E2+-+4cot+%2B+sqrt%283%29=+0+
Then find cot^2 + tan^2
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%283%29%28cot%28theta%29%29%5E2+-+4cot%28theta%29+%2B+sqrt%283%29=+0+
Let's make a change of varaible so I do not have to write cot%28theta%29 so many times.
cot%28theta%29=y
Now I can write the equation as
sqrt%283%29y%5E2-4y%2Bsqrt%283%29=0
We have to solve that equation
It may not be obvious, but we can solve that equation by factoring.
The left diside of the equal sign can be factored to get
%28sqrt%283%29y-1%29%28y-sqrt%283%29%29=0
That leads to 2 solutions
sqrt%283%29y-1=0<-->y=1%2Fsqrt%283%29=sqrt%283%29%2F3
y-sqrt%283%29=0<-->y=sqrt%283%29
Since cot%28theta%29=1%2Ftan%28theta%29, the solutions look very symmetrical:
cot%28theta%29=1%2Fsqrt%283%29-->tan%28theta%29=1%2Fcot%28theta%29=sqrt%283%29
and
cot%28theta%29=sqrt%283%29-->tan%28theta%29=1%2Fcot%28theta%29=1%2Fsqrt%283%29
One function is sqrt%283%29 and the other is 1%2Fsqrt%283%29=sqrt%283%29%2F3.
For either set of solutions,


PS _ No, I would not have tried to use the quadratic formula to solve
sqrt%283%29y%5E2-4y%2Bsqrt%283%29=0, but I did not think of the factorization right away.
The way I really figured out the solutions was dividing both sides by sqrt%283%29 (same as multiplying times sqrt%283%29%2F3 if you like it better that way, and then "completing the square:
sqrt%283%29y%5E2-4y%2Bsqrt%283%29=0
y%5E2-%284%2Fsqrt%283%29%29y%2B1=0
y%5E2-2%282%2Fsqrt%283%29%29y=-1
y%5E2-2%282%2Fsqrt%283%29%29y%2B%282%2Fsqrt%283%29%29%5E2=-1%2B%282%2Fsqrt%283%29%29%5E2
%28y-2%2Fsqrt%283%29%29%5E2=-1%2B4%2F3
%28y-2%2Fsqrt%283%29%29%5E2=1%2F3
Then, taking square roots:
--> --> system%28y=3%2Fsqrt%283%29%2C%22or%22%2Cy=1%2Fsqrt%283%29%29

EXTRA:
The problem did not ask about the angles, but the angles in the first quadrant that have those tangent and cotangent values are pi%2F6 and pi%2F3 (or 30%5Eo and 60%5Eo if you prefer degrees).