Please answer the following questions ,though it's in one thread , please give answer one by one please

Dear Student, 27:We know that sin2A+cos2A= 1 and x2-y2= x-yx+yhere sin4θ-cos4θ+1cosec2θ= sin2θ2-cos2θ2+1cosec2θ=sin2θ-cos2θsin2θ+cos2θ+1cosec2θ= sin2θ-cos2θ1+1cosec2θ               : using sin2θ+cos2θ=1=sin2θ-cos2θ+sin2θ+cos2θcosec2θ    : using 1 =sin2θ+cos2θ = 2sin2θ×cosec2θ =2sin2θ×1sin2θ=2Please post single query in a threadRegards

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