Cos 12

$cos^2x+sin^2x=1<=> h^2 + sin^212=1<=>sin12= \sqrt{1-h^2}=\sqrt{(1-h)(1+h)}$

$cot12=\frac{cos12}{sin12}=\frac{h}{\sqrt{(1-h)(1+h)}}$

By the Pythagorean Identity,

$\sin^2(12^\circ)+\cos^2(12^\circ)\ =\ 1$

                                                      We are given that   cos(12°) = h   so we can substitute  h  in for  cos(12°)

$\sin^2(12^\circ)+h^2\ =\ 1$

                                                      Subtract  h²  from both sides of the equation.

$\sin^2(12^\circ)\ =\ 1-h^2$

                                                      Because  12°  is in Quadrant I,  sin(12°)  is positive. So take positive sqrt of both sides.

$\sin(12^\circ)\ =\ \sqrt{1-h^2}$

By definition of cotangent,

$\cot(12^\circ)\ =\ \frac{\cos(12^\circ)}{\sin(12^\circ)}$

                                          Substitute  h  in for  cos(12°)  and substitute  √[ 1 - h² ]  in for  sin(12°)

$\cot(12^\circ)\ =\ \frac{h}{\sqrt{1-h^2}}$_