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Prove that :-tan³A/1+tan²A + cot³A/1+cot²A = …

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Prove that :-tan³A/1+tan²A + cot³A/1+cot²A = secA cosecA -2sinAcosA Anyone can answer my question Anyone have power ?✊?✊????????

    • Posted by Nikita Singh 6 years, 5 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Himanshi Sharma 6 years, 5 months ago

    L.H.S=tan^3A/(1+tan^2A)+cot^3A/(1+cot^2A) =tan^3A/sec^2A+cot^3A/cosec^2A =(sin^3A/cos^3A)/(1/cos^2A)+(cos^3A/sin^A)/(1/sin^2A) =sin^3A/cosA+cos^3A/sinA =(sin^4A+cos^A)/sinA.cosA =[(sin^2A+cos^2A)^2-2sin^2A.cos^2A]/sinA.cosA =(1-2sin^A.cos^A)/sinA.cosA R.H.S=secA.cosecA-2sinA.cosA =1/cosA.1/sinA-2sinAcosA =(1-sin^2A.cos^2A)/sinA.cosA Hence,L.H.S=R.H.S.......

    3Thank You

    Aastha Dangi 6 years, 5 months ago

    LHS = Tan^3A / ( 1+ Tan^2A) + Cot^3A / (1 + Cot^2a) = Tan^3A / Sec^2A + Cot^3A / 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 = [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA = ( 1- 2Sin^A.Cos^A)/ SinA.CosA RHS = SecA CosecA - 2sinAcosA = 1/CosA . 1/SinA - 2SinACosA = (1 - Sin^2A.Cos^2A) / sinAcosA Hence LHS = RHS (PROVED)

    3Thank You

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