If C= 2 pi by 9,prove that root3.cosec C-secC

Question

It is based on triangle ABC where A=2 pi by 9 and B = 2 picture by 3

Answer 1

2π/9+2π/3=2π(1/9+3/9)=8π/9. So angle C=π-8π/9=π/9 (because the angles of a triangle add up to 180 degrees=π radians). cosecC=2.9238 approx; secC=1.0642 approx. √3=1.7321 approx.

The question doesn't provide an equation to prove, because only an expression is given.

The angles of the triangle are C=π/9,  A=2π/9 and B=6π/9=2π/3.

If x=π/9 then 3x=π/3 and

cosπ/3=1/2=cos(3x)=cos(2x)cos(x)-sin(2x)sin(x)=(1-2sin^2(x))cos(x)-2sin^2(x)cos(x).

cos(x)(1-4sin^2(x))=1/2.

sinπ/3=√3/2=sin(3x)=sin(2x)cos(x)+cos(2x)sin(x)=2sin(x)cos^2(x)+(2cos^2(x)-1)sin(x).

sin(x)(4cos^2(x)-1)=√3/2; √3cosec(x)=2(4cos^2(x)-1)=2(2cos(x)-1)(2cos(x)+1).