Prove that CotA/2-3cot3A/2=4sinA/1+2cosA?

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Answer 1 · 2018-02-11T19:42:57

Please refer to a Proof in the Explanation.

We know that, $tan 3 x = \frac{3 tan x - {tan}^{3} x}{1 - 3 {tan}^{2} x}$ $tan3x=(3tanx-tan^3x)/(1-3tan^2x)$ .

$:. cot3x=1/(tan3x)=(1-3tan^2x)/(3tanx-tan^3x)$ .

Let, $tan(A/2)=t$ . Then,

$cot(A/2)-3cot(3A/2)=1/t-(3(1-3t^2))/(3t-t^3)$ ,

$=1/t-(3(1-3t^2))/(t(3-t^2)$ ,

$={(3-t^2)-(3-9t^2)}/(t(3-t^2)$ ,

$=(8t)/(3-t^2)$ ,

$={(8t)/(1+t^2)}-:{(3-t^2)/(1+t^2)}$ ,

$={(8t)/(1+t^2)}-:{{(1+t^2)+(2-2t^2)}/(1+t^2)}$ ,

$={(8t)/(1+t^2)}-:{(1+t^2)/(1+t^2)+(2(1-t^2))/(1+t^2)}$ ,

$={4((2t)/(1+t^2))}-:{1+2((1-t^2)/(1+t^2))}$ .

Since, $sinA=(2tan(A/2))/(1+tan^2(A/2)), and, cos2y=(1-tan^2(A/2))/(1+tan^2(A/2))$ ,

$cot(A/2)-3cot(3A/2)=4sinA-:(1+2cosA)$ , as desired!