cot inverse [ root ( 1 + sinx ) + root ( 1 - sinx ) / root ( 1 + sinx ) - root ( 1 - sinx ) ] = x/2

Hi Manonarayanan,
Please find below the solution to the asked query:

L.H.S.=cot-11+sin x+1-sin x1+sin x-1-sin xcos2α2+sin2α2=1 and sin α=2sinα2.cosα2, we getL.H.S.=cot-1cos2x2+sin2x2+2sinx2.cosx2+cos2x2+sin2x2-2sinx2.cosx2cos2x2+sin2x2+2sinx2.cosx2-cos2x2+sin2x2-2sinx2.cosx2Using identity a±b2=a2+b2±2ab , we get,L.H.S.=cot-1cosx2+sinx22+cosx2-sinx22cosx2+sinx22-cosx2-sinx22=cot-1cosx2+sinx2+cosx2-sinx2cosx2+sinx2-cosx2-sinx2=cot-1cosx2+sinx2+cosx2-sinx2cosx2+sinx2-cosx2+sinx2=cot-12cosx22sinx2=cot-1cosx2sinx2=cot-1cot x2      Ascos αsin α=cot αL.H.S.=x2=R.H.S. Hence proved

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