Question

What is `cot(theta/2)` in terms of `csctheta`?

Answer 1

It is

`cot²(x) + 1 = csc²(x)`

hence for `x=theta/2`

`cot(theta/2)=sqrt(csc^2(theta/2)-1)`

But

#csc(theta/2)=1/(sin(theta/2))=1/(sin(theta)/[2*cos(theta/2)])= 2*cos(theta/2)/(sin(theta))#

But #cos(2theta)=2cos^2theta-1=> cos^2theta=1/2*(1+cos(2theta))#

Hence #cos(theta/2)=1/sqrt2*(1+costheta)=1/sqrt2*[1+sqrt(1-sintheta)]=> cos(theta/2)=1/sqrt2*[1+sqrt(1-1/(csctheta))]#

And

#csc(theta/2)=2*cos(theta/2)/(sin(theta))=> csc(theta/2)=2*[1/sqrt2*[1+sqrt(1-1/(csctheta))]]*csctheta=> csc(theta/2)=sqrt2*csctheta*[1+sqrt(1-1/(csctheta))]#

Finally

`cot(theta/2)=sqrt([sqrt2*csctheta*(1+sqrt(1-1/(csctheta)))]^2-1)`