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Question
Tan3 A/ 1+tan2A + cot3A / 1+ cot2A = secA cosecA - 2sinA cosA
Tan3 A/ 1+tan2A + cot3A / 1+ cot2A = secA cosecA - 2sinA cosA
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Solution
LHS = Tan^3A / ( 1+ Tan^2A) + Cot^3A / (1 + Cot^2A)
= Tan^3A / Sec^2A + Cot^3A / Cosec^2A [ 1+ Tan^2A=Sec^2A & 1 + Cot^2A=Cosec^2A ]
= (sin^3A/cos^3A) / (1 / Cos^2A) + (Cos^3A/Sin^3A) / (1 / Sin^2A)
= Sin^3A/CosA + Cos^3A/SinA
= (Sin^4A + Cos^4A) / SinA.CosA [multiplying SinA/SinA in 1st term & CosA/CosA in 2nd]
= [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA [a^2 + b^2 = (a+b)^2 -2ab]
= ( 1- 2Sin^A.Cos^A)/ SinA.CosA
RHS = SecA CosecA - 2sinAcosA
=( 1/CosA . 1/SinA) - 2SinACosA
=(1/(CosA.SinA)) - 2SinACosA
= (1 - 2Sin^2A.Cos^2A) / sinAcosA
Hence LHS = RHS (PROVED)
= Tan^3A / Sec^2A + Cot^3A / Cosec^2A [ 1+ Tan^2A=Sec^2A & 1 + Cot^2A=Cosec^2A ]
= (sin^3A/cos^3A) / (1 / Cos^2A) + (Cos^3A/Sin^3A) / (1 / Sin^2A)
= Sin^3A/CosA + Cos^3A/SinA
= (Sin^4A + Cos^4A) / SinA.CosA [multiplying SinA/SinA in 1st term & CosA/CosA in 2nd]
= [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA [a^2 + b^2 = (a+b)^2 -2ab]
= ( 1- 2Sin^A.Cos^A)/ SinA.CosA
RHS = SecA CosecA - 2sinAcosA
=( 1/CosA . 1/SinA) - 2SinACosA
=(1/(CosA.SinA)) - 2SinACosA
= (1 - 2Sin^2A.Cos^2A) / sinAcosA
Hence LHS = RHS (PROVED)
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