Question

sin2A. tanA+cos2A.cotA+2sinA.cosA=tanA.cotA​

Answer 1

Answer:

LHS

= Sin²A.TanA  + Cos²A.CotA  + 2SinA.CosA

using TanA = SinA/CosA    & CotA = CosA/SinA

= Sin²A.SinA/CosA  + Cos²A.CosA/SinA  + 2SinA.CosA

= Sin³A/CosA  + Cos³A./SinA  + 2SinA.CosA

= ( Sin⁴A  + Cos⁴A  + 2Sin²A.Cos²A )/CosASinA

= ( (Sin²A)²  + (Cos²A)²  + 2Sin²A.Cos²A )/CosASinA

using a² + b² + 2ab = (a + b)²

a = Sin²A  & b = Cos²A

= ((Sin²A)  + (Cos²A))²/CosA.SinA

using Sin²A + Cos²A = 1

= 1/CosA.SinA

= (Sin²A + Cos²A)/CosASinA

= Sin²A/CosASinA + Cos²A/CosASinA

= SinA/CosA  + CosA/SinA

= TanA  + CotA

= RHS

QED

Proved

Step-by-step explanation: