cot^-1[(root1-sinx + root1+sinx) / (root1-sinx - root1+sinx)] [where x belongs to (0,pi+2)] Share with your friends Share 0 Ankita Agarwal answered this Dear Student , Given : Let y=cot-11-sin x+1+sin x1-sin x-1+sin x Then using rationalization by denominator=cot-11-sin x+1+sin x1-sin x-1+sin x ×1-sin x+1+sin x1-sin x+1+sin x =cot-11-sin x+1+sin x21-sin x-1+sin x 1-sin x+1+sin x using a2-b2=a+ba-b and a+b2=a2+b2+2ab=cot-11-sin x+1+sin x-21-sin x1+sin x1+sin x-1-sinx=cot-12+21-sin2x2sin x=cot-11+cosxsinx using sin2+cos2x=1=cot-12cos2x22sinx2cosx2 using sin2A=2sinAcosA and cos 2A=2cos2A-1=cot-1cotx2=x2 Regards 0 View Full Answer
