document.write( "Question 789080: Please help me solve this problem:
\n" ); document.write( "If $\"+sqrt%283%29cot%5E2+-+4cot+%2B+sqrt%283%29=+0+\"$
\n" ); document.write( "Then find cot^2 + tan^2 \n" ); document.write( "

Algebra.Com's Answer #478738 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"+sqrt%283%29%28cot%28theta%29%29%5E2+-+4cot%28theta%29+%2B+sqrt%283%29=+0+\"$
\n" ); document.write( "Let's make a change of varaible so I do not have to write $\"cot%28theta%29\"$ so many times.
\n" ); document.write( " $\"cot%28theta%29=y\"$
\n" ); document.write( "Now I can write the equation as
\n" ); document.write( " $\"sqrt%283%29y%5E2-4y%2Bsqrt%283%29=0\"$
\n" ); document.write( "We have to solve that equation
\n" ); document.write( "It may not be obvious, but we can solve that equation by factoring.
\n" ); document.write( "The left diside of the equal sign can be factored to get
\n" ); document.write( " $\"%28sqrt%283%29y-1%29%28y-sqrt%283%29%29=0\"$
\n" ); document.write( "That leads to 2 solutions
\n" ); document.write( " $\"sqrt%283%29y-1=0\"$ <--> $\"y=1%2Fsqrt%283%29=sqrt%283%29%2F3\"$
\n" ); document.write( " $\"y-sqrt%283%29=0\"$ <--> $\"y=sqrt%283%29\"$
\n" ); document.write( "Since $\"cot%28theta%29=1%2Ftan%28theta%29\"$ , the solutions look very symmetrical:
\n" ); document.write( " $\"cot%28theta%29=1%2Fsqrt%283%29\"$ --> $\"tan%28theta%29=1%2Fcot%28theta%29=sqrt%283%29\"$
\n" ); document.write( "and
\n" ); document.write( " $\"cot%28theta%29=sqrt%283%29\"$ --> $\"tan%28theta%29=1%2Fcot%28theta%29=1%2Fsqrt%283%29\"$
\n" ); document.write( "One function is $\"sqrt%283%29\"$ and the other is $\"1%2Fsqrt%283%29=sqrt%283%29%2F3\"$ .
\n" ); document.write( "For either set of solutions,
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\n" ); document.write( "PS _ No, I would not have tried to use the quadratic formula to solve
\n" ); document.write( " $\"sqrt%283%29y%5E2-4y%2Bsqrt%283%29=0\"$ , but I did not think of the factorization right away.
\n" ); document.write( "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:
\n" ); document.write( " $\"sqrt%283%29y%5E2-4y%2Bsqrt%283%29=0\"$
\n" ); document.write( " $\"y%5E2-%284%2Fsqrt%283%29%29y%2B1=0\"$
\n" ); document.write( " $\"y%5E2-2%282%2Fsqrt%283%29%29y=-1\"$
\n" ); document.write( " $\"y%5E2-2%282%2Fsqrt%283%29%29y%2B%282%2Fsqrt%283%29%29%5E2=-1%2B%282%2Fsqrt%283%29%29%5E2\"$
\n" ); document.write( " $\"%28y-2%2Fsqrt%283%29%29%5E2=-1%2B4%2F3\"$
\n" ); document.write( " $\"%28y-2%2Fsqrt%283%29%29%5E2=1%2F3\"$
\n" ); document.write( "Then, taking square roots:
\n" ); document.write( "

-->

--> $\"system%28y=3%2Fsqrt%283%29%2C%22or%22%2Cy=1%2Fsqrt%283%29%29\"$
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\n" ); document.write( "EXTRA:
\n" ); document.write( "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). \n" ); document.write( "

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