prove that tanx.tan(x+60)+tanx.tan(x-60)+tan(x+60).tan(x-60)

1.	prove that tanx.tan(x+60)+tanx.tan(x-60)+tan(x+60).tan(x-60)
Answer» \xa0Tan(A-B) =TanA-TanB/(1+TanATanB) => TanATanB+1=(TanA-TanB)/Tan(A-B)SoTanxTan(60+x)+1 ={Tan(60+x)-Tan60}/Tan60 -----1Tanxtan(x-60)+1= {Tanx- Tan(60-x)}/tan60---------2Tan(x+60)tan(x-60)+1 ={tan(60+x)-tan(x-60)}/tan120--------3Adding 1,2 and 3we get TanxTan(60+x)+1+Tanxtan(x-60)+1+Tan(x+60)tan(x-60)+1=0soTanx tan(x+60)+tanx tan(x-60)+tan(x+60) tan(x-60)=-3

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