If cosecx−sinx=a3andsecx−cosx=b3;then what is the value ofa2b2(a2+b2)−
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a3=cosecx−sinxandb3=secx−cosx(given)⇒a3=1/sinx−sinx andb3=1/cosx−cosx⇒a3=(1−sin2x)/sinx andb3=(1−cos2x)/cosx⇒a3=(cos2x/sinx)andb3=(sin2x/cosx)⇒a=3√(cos2x/sinx) andb=3√(sin2x/cosx)Now the given expression−a2b2(a2+b2)=a4b2+a2b4
Now by putting the values of a and b in it then the expression=[(cos2x/sinx)(4/3)][(sin2x/cosx)(2/3)]+[(cos2x/sinx)(2/3)(sin2x/cosx)(4/3)]=cos2x+sin2x=1