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Question

Tan​​​​3 A/ 1+tan​​​​2​​​A + cot​3​​​A / 1+ cot​2​​​A = 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)


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