Prove that $cot(A/2)-3cot((3A)/2)=(4sinA)/(1+2cosA)$ ?

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Answer 1 · 2018-05-24T23:22:41

$LHS=cot(A/2)-3cot((3A)/2)$

$=cos(A/2)/sin(A/2)-(3cos((3A)/2))/sin((3A)/2)$

$=(cos(A/2)sin((3A)/2)-3cos((3A)/2)sin(A/2))/(sin(A/2)sin((3A)/2))$

$=(cos(A/2)sin((3A)/2)-cos((3A)/2)sin(A/2)-2cos((3A)/2)sin(A/2))/(sin(A/2)sin((3A)/2))$

$=(sin((3A)/2-A/2)-(sin((3A)/2+A/2)-sin((3A)/2-A/2)))/(sin(A/2)sin((3A)/2))$

$=(sinA-(sin2A-sinA))/(sin(A/2)sin((3A)/2))$

$=(2sinA-sin2A)/(sin(A/2)sin((3A)/2))$

$=(2sinA-2sinAcosA)/(sin(A/2)sin((3A)/2))$

$=(4sinA(1-cosA))/(2sin(A/2)sin((3A)/2))$

$=(4sinA(1-cosA))/(cos((3A)/2-A/2)-cos((3A)/2+A/2))$

$=(4sinA(1-cosA))/(cosA-cos2A)$

$=(4sinA(1-cosA))/(cosA-(2cos^2A-1))$

$=(4sinA(1-cosA))/(1+cosA-2cos^2A)$

$=(4sinA(1-cosA))/(1+2cosA-cosA-2cos^2A)$

$=(4sinA(1-cosA))/((1+2cosA)-cosA(1+2cosA))$

$=(4sinA(1-cosA))/((1+2cosA)(1-cosA))$

$=(4sinA)/(1+2cosA)=RHS$

Prove that cot(A/2)-3cot((3A)/2)=(4sinA)/(1+2cosA)cot(A/2)-3cot((3A)/2)=(4sinA)/(1+2cosA)?