Question 1011732
{{{cot(x)cot(2x)}}}{{{""=""}}}{{{1}}}
<pre>
We normally aren't given double angle formulas for cotangents.
We usually only are given double angle formulas for tangents.

Since we know that {{{cot(theta)=1/tan(theta)}}}, let's
convert to tangents:

{{{cot(x)cot(2x)}}}{{{""=""}}}{{{1}}}

{{{(1/tan(x))(1/tan(2x))}}}{{{""=""}}}{{{1}}}

{{{1/(tan(x)tan(2x))}}}{{{""=""}}}{{{1}}}

We take reciprocals of both sides, and since the reciprocal
of 1 is 1, we have

{{{tan(x)tan(2x)}}}{{{""=""}}}{{{1}}}

We use the double-angle formula that we are always given:
which is {{{tan(2theta)=(2tan(theta))/(1-tan^2(theta))}}}


{{{tan(x)((2tan(x))/(1-tan^2(x)))}}}{{{""=""}}}{{{1}}}

{{{(tan(x)/1^"")((2tan(x))/(1-tan^2(x)))}}}{{{""=""}}}{{{1}}}

{{{(2tan^2(x))/(1-tan^2(x))}}}{{{""=""}}}{{{1}}}

Multiply both sides by the denominator on the left:

{{{2tan^2(x)}}}{{{""=""}}}{{{1-tan^2(x)}}}

Solve for tan<sup>2</sup>(x)

{{{3tan^2(x)}}}{{{""=""}}}{{{1}}}

{{{3tan^2(x)}}}{{{""=""}}}{{{1}}}

Take square roots 

{{{tan(x)}}}{{{""=""}}}{{{"" +- sqrt(1/3)}}}

{{{tan(x)}}}{{{""=""}}}{{{"" +- 1/sqrt(3)}}}

Remembering the special 30°-60°-90° right triangle:

{{{drawing(100,30000/463,-.2,1.832,-.1,1.1,locate(1.1,.3,sqrt(3)),
locate(.35,.26,"30°"),locate(1.75,.65,1),red(arc(0,0,1.9,-1.9,0,30)),
locate(.7,.7,2),
triangle(0,0,sqrt(3),0,sqrt(3),1) )}}}

we know by the ± that the angle can be in any quadrant
with a 30° reference angle, so the solutions are

30°, 150°, 210°, 330°  plus any integer n times 360°.

or in radians

{{{pi/6}}},{{{5pi/6}}},{{{7pi/6}}},{{{11pi/6}}} plus any integer n times {{{2pi}}}.

Edwin</pre>