Sunday, 4 September 2016

f Cosec-Sin=a3 and Sec-Cos=b3, then a2b2(a2+b2) =?

If Cosec-Sin=a3 and Sec-Cos=b3, then a2b2(a2+b2) =?
Answer: Cosec-Sin=a3 and Sec-Cos=b3, then a2b2(a2+b2)
=> Cosec-Sin=a3=> a3=Cos2/sin and Sec-Cos=b3=>b3= Sin2/cos
=>a2b2(a2+b2) =a4b2+a2b4= (a2b+ab2)2-2a3b3=[(a6b3)1/3+(a3b6)1/3]2-2a3b3

=>Applying the values of a3=Cos2/sin and b3= Sin2/cos in the expression [(a6b3)1/3+(a3b6)1/3]2-2a3b3, [(a6b3)1/3+(a3b6)1/3]2-2a3b3=(Sin+Cos)2- 2Sin.Cos=1(Answer)
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